Demystifying the Natural Logarithm (ln)

After understanding the exponential function, our next target is the natural logarithm.

Given how the natural log is described in math books, there’s little “natural” about it: it’s defined as the inverse of e^x, a strange enough exponent already.

But there’s a fresh, intuitive explanation: The natural log gives you the time needed to reach a certain level of growth.

Suppose you have an investment in gummy bears (who doesn’t?) with an interest rate of 100% per year, growing continuously. If you want 10x growth, assuming continuous compounding, you’d wait only ln(10) or 2.302 years. Don’t see why it only takes a few years to get 10x growth? Don’t see why the pattern is not 1, 2, 4, 8? Read more about e.

e and the Natural Log are twins:

  • e^x is the amount of continuous growth after a certain amount of time.
  • Natural Log (ln) is the amount of time needed to reach a certain level of continuous growth

Not too bad, right? While the mathematicians scramble to give you the long, technical explanation, let’s dive into the intuitive one.

E is About Growth

The number e is about continuous growth. As we saw last time, e^x lets us merge rate and time: 3 years at 100% growth is the same as 1 year at 300% growth, when continuously compounded.

We can take any combination of rate and time (50% for 4 years) and convert the rate to 100% for convenience (giving us 100% for 2 years). By converting to a rate of 100%, we only have to think about the time component:

\displaystyle{e^x = e^{rate \cdot time} = e^{1.0 \cdot time} = e^{time}}

Intuitively, e^x means:

  • How much growth do I get after after x units of time (and 100% continuous growth)
  • For example: after 3 time periods I have e^3 = 20.08 times the amount of “stuff”.

e^x is a scaling factor, showing us how much growth we’d get after x units of time.

Natural Log is About Time

The natural log is the inverse of e, a fancy term for opposite. Speaking of fancy, the Latin name is logarithmus naturali, giving the abbreviation ln.

Now what does this inverse or opposite stuff mean?

  • e^x lets us plug in time and get growth.
  • ln(x) lets us plug in growth and get the time it would take.

For example:

  • e^3 is 20.08. After 3 units of time, we end up with 20.08 times what we started with.
  • ln(20.08) is about 3. If we want growth of 20.08, we’d wait 3 units of time (again, assuming a 100% continuous growth rate).

With me? The natural log gives us the time needed to hit our desired growth.

Logarithmic Arithmetic Is Not Normal

You’ve studied logs before, and they were strange beasts. How’d they turn multiplication into addition? Division into subtraction? Let’s see.

What is ln(1)? Intuitively, the question is: How long do I wait to get 1x my current amount?

Zero. Zip. Nada. You’re already at 1x your current amount! It doesn’t take any time to grow from 1 to 1.

  • ln(1) = 0

Ok, how about a fractional value? How long to get 1/2 my current amount? Assuming you are growing continuously at 100%, we know that ln(2) is the amount of time to double. If we reverse it (i.e., take the negative time) we’d have half of our current value.

  • ln(.5) = – ln(2) = -.693

Makes sense, right? If we go backwards (negative time) .693 seconds we’d have half our current amount. In general, you can flip the fraction and take the negative: ln(1/3) = – ln(3) = -1.09. This means if we go back 1.09 units of time, we’d have a third of what we have now.

Ok, how about the natural log of a negative number? How much time does it take to “grow” your bacteria colony from 1 to -3?

It’s impossible! You can’t have a “negative” amount of bacteria, can you? At most (er… least) you can have zero, but there’s no way to have a negative amount of the little critters. Negative bacteria just doesn’t make sense.

  • ln(negative number) = undefined

Undefined just means “there is no amount of time you can wait” to get a negative amount.

Logarithmic Multiplication is Mighty Fun

How long does it take to grow 4x your current amount? Sure, we could just use ln(4). But that’s too easy, let’s be different.

We can consider 4x growth as doubling (taking ln(2) units of time) and then doubling again (taking another ln(2) units of time):

  • Time to grow 4x = ln(4) = Time to double and double again = ln(2) + ln(2)

Interesting. Any growth number, like 20, can be considered 2x growth followed by 10x growth. Or 4x growth followed by 5x growth. Or 3x growth followed by 6.666x growth. See the pattern?

  • ln(a*b) = ln(a) + ln(b)

The log of a times b = log(a) + log(b). This relationship makes sense when you think in terms of time to grow.

If we want to grow 30x, we can wait ln(30) all at once, or simply wait ln(3), to triple, then wait ln(10), to grow 10x again. The net effect is the same, so the net time should be the same too (and it is).

How about division? ln(5/3) means: How long does it take to grow 5 times and then take 1/3 of that?

Well, growing 5 times is ln(5). Growing 1/3 is -ln(3) units of time. So

  • ln(5/3) = ln(5) – ln(3)

Which says: Grow 5 times and “go back in time” until you have a third of that amount, so you’re left with 5/3 growth. In general we have

  • ln(a/b) = ln(a) – ln(b)

I hope the strange math of logarithms is starting to make sense: multiplication of growth becomes addition of time, division of growth becomes subtraction of time. Don’t memorize the rules, understand them.

Using Natural Logs With Any Rate

“Sure,” you say, “This log stuff works for 100% growth but what about the 5% I normally get?”

It’s no problem. The “time” we get back from ln() is actually a combination of rate and time, the “x” from our e^x equation. We just assume 100% to make it simple, but we can use other numbers.

Suppose we want 30x growth: plug in ln(30) and get 3.4. This means:

  • e^x = growth
  • e^3.4 = 30

And intuitively this equation means “100% return for 3.4 years is 30x growth”. We can consider the equation to be:

\displaystyle{e^{x} = e^{rate \cdot time}}

\displaystyle{e^{100 \% \cdot 3.4 years} = 30}

We can modify “rate” and “time”, as long as rate * time = 3.4. For example, suppose we want 30x growth — how long do we wait assuming 5% return?

  • ln(30) = 3.4
  • rate * time = 3.4
  • .05 * time = 3.4
  • time = 3.4 / .05 = 68 years

Intuitively, I think “ln(30) = 3.4, so at 100% growth it will take 3.4 years. If I double the rate of growth, I halve the time needed.”

  • 100% for 3.4 years = 1.0 * 3.4 = 3.4
  • 200% for 1.7 years = 2.0 * 1.7 = 3.4 [200% growth means half the time]
  • 50% for 6.8 years = 0.5 * 6.8 = 3.4 [50% growth means double the time]
  • 5% for 68 years = .05 * 68 = 3.4 [5% growth means 20x the time]

Cool, eh? The natural log can be used with any interest rate or time as long as their product is the same. You can wiggle the variables all you want.

Awesome example: The Rule of 72

The Rule of 72 is a mental math shortcut to estimate the time needed to double your money. We’re going to derive it (yay!) and even better, we’re going to understand it intuitively.

How long does it take to double your money at 100% interest, compounded every year?

Uh oh. We’ve been using natural log for continuous rates, but now you’re asking for yearly interest? Won’t this mess up our formula? Yes, it will, but at reasonable interest rates like 5%, 6% or even 15%, there isn’t much difference between yearly compounded and fully continuous interest. So the rough formula works, uh, roughly and we’ll pretend we’re getting fully continuous interest.

Now the question is easy: How long to double at 100% interest? ln(2) = .693. It takes .693 units of time (years, in this case) to double your money with continuous compounding with a rate of 100%.

Ok, what if our interest isn’t 100% What if it’s 5% or 10%?

Simple. As long as rate * time = .693, we’ll double our money:

  • rate * time = .693
  • time = .693/rate

So, if we only had 10% growth, it’d take .693 / .10 or 6.93 years to double.

To simplify things, let’s multiply by 100 so we can talk about 10 rather than .10:

  • time to double = 69.3/rate, where rate is assumed to be in percent.

Now the time to double at 5% growth is 69.3/5 or 13.86 years. However, 69.3 isn’t the most divisible number. Let’s pick a close neighbor, 72, which can be divided by 2, 3, 4, 6, 8 and many more numbers.

  • time to double = 72/rate

which is the rule of 72! Easy breezy.

If you want to find the time to triple, you’d use ln(3) ~ 109.8 and get

  • time to triple = 110 / rate

Which is another useful rule of thumb. The Rule of 72 is useful for interest rates, population growth, bacteria cultures, and anything that grows exponentially.

Where to from here?

I hope the natural log makes more sense — it tells you the time needed for any amount of exponential growth. I consider it “natural” because e is the universal rate of growth, so ln could be considered the “universal” way to figure out how long things take to grow.

When you see ln(x), just think “the amount of time to grow to x”. In an upcoming article I’ll bring e and ln together, and the sweet aroma of math will fill the air.

Appendix: The Natural Log of E

Quick quiz: What’s ln(e)?

  • The math robot says: Because they are defined to be inverse functions, clearly ln(e) = 1
  • The intuitive human: ln(e) is the amount of time it takes to get “e” units of growth (about 2.718). But e is the amount of growth after 1 unit of time, so ln(e) = 1.

Think intuitively.

Other Posts In This Series

  1. A Visual, Intuitive Guide to Imaginary Numbers
  2. Surprising Uses of the Pythagorean Theorem
  3. Rescaling the Pythagorean Theorem
  4. Intuitive Guide to Angles, Degrees and Radians
  5. Intuitive Understanding Of Euler's Formula
  6. Intuitive Understanding of Sine Waves
  7. An Interactive Guide To The Fourier Transform
  8. Intuitive Arithmetic With Complex Numbers
  9. Understanding Why Complex Multiplication Works
  10. An Intuitive Guide To Exponential Functions & e
  11. Demystifying the Natural Logarithm (ln)
  12. Understanding Exponents (Why does 0^0 = 1?)
  13. A Visual Guide to Simple, Compound and Continuous Interest Rates
  14. Using Logarithms in the Real World
  15. How To Measure Any Distance With The Pythagorean Theorem
Kalid Azad loves those Aha! moments when an idea finally clicks. BetterExplained is dedicated to learning with intuition, not blind memorization, and is honored to serve 250k readers each month.

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234 Comments

  1. Uhm, wouldn’t bacteria grown at 100% per day take more than 3 days? first day gives you 2x growth, second gives you 4x, third gives you 8x, sometime after 3rd but before 4th gives you 10x. Perhaps .3 days, dunno.
    Sorry, I couldn’t read past that seemingly huge error.

  2. Ah, that’s the magic of continuous compound growth. Consider how much bacteria you have at the 12-hour mark: 1.5 (50% more than you started).

    After another 12 hours that amount (1.5) has time to grow another 50%: 1.5 * 1.5 = 2.25, even better than double at the end of the day.

    Now take a smaller interval, like 6-hours. After 6 hours you have 25% growth: 1.25

    and after 4 periods of 25% growth you have 1.25^4 = 2.44 times your original.

    If you keep taking smaller and smaller time intervals, you get a net compound growth rate of 2.71828 per day, which is e (take a look at the article on e for more details). I’ll amend the article to clarify.

  3. Perhaps, but that’s NOT what you said, you said they had 100% growth in 1 day. You didn’t say 125% growth in 1 day.

  4. The example you’re giving seems to be based on compounding interest; if you have a 5% interest rate (yearly) BUT you compound it continously you get an APR or whatever of 5.25% or whatever. Basically it works out that ln would be useful IF you wanted to take something that didn’t compound continously, and see what would happen if it did. By definition, a 100% increase in a colony of bacteria in 24 hours means that it will actually have double in size in 24 hours. Yes it would have been growing continously, but the end result would be 100% growth, not e^1.

  5. That’s a good point, I think I’ll rephrase the bacteria example to use money, where the concepts of “simple” and “compound” interest are more natural.

    You can imagine bacteria that grows at 100% “simple interest”, and the net result would be over 100% growth if the mini-bacteria which are made after a half-day create new bacteria of their own. But describing it as 100% continuous growth, as worded, may be confusing.

    The meta-point, which I’m pretty sure we both agree on, is that ln can give you the time something growing continuously (e^x) would take.

    And the cool thing is that ln can even help figure out non-continuous growth (2^x, like the normal bacteria case) as well. I’ll be writing more on this.

  6. I’m not sure that I agree on the “something” grows part. It doesn’t really seem to apply to anything more than interest, or something I can’t think of that might not grow in cleanly divisible increments. As for the ln being useful for the bacteria growth, actually not all. That would be Log base 2 rather than Log base e.

  7. Uh oh, you’re going to make me give away the crown secret: e and ln can be used in a change-of-base formula, so you actually don’t need a separate log base 2 to get the expected number of doublings :) (It’s convenient to have a separate log base 2, but not strictly needed).

  8. Wow… i’ve been casually using ln(N) in the context of “big-O notation” in programming for years, but this article is the first to really get my on my way to being able to USE the function. i first tried the related Wikipedia articles and, like you mentioned, was only mystified by the circular references between ln and ‘e’.

  9. Hey Stephan, thanks for the mail! There’s so many topics that we just take for granted and use mechanically, it’s a lot of fun to really understand them. E and natural log was like this for a long time for me, too.

  10. Hi my friend… congratulations for an amazing site! This is the way these things should be taught to children! Will be spreading the word to as many people as I can! Since you’re into e, I wonder if you could write something demystifying a bit hyperbolic trigonometry.

  11. Thanks John, I’m happy you liked the site! I think hyperbolic geometry would be a good topic — personally, I need to investigate it a bit more! But I’d love to share what I learn when I do.

  12. ln(a*b) = ln(a) + ln(b)

    The log of a times b = log a + log b. This relationship makes sense when you think about it being the time to grow.

    If we want to grow 30x, we can wait ln(30) all at once, or simply wait ln(3), to triple, then ln(10), to grow 10x again. The net effect is the same, so the net time should be the same too (and it is).

    this is very well said;
    i don’t know that i’ve ever
    seen it put in quite this way.

    on the other hand, you’ve confused
    the issue mightily by referring
    to “simple interest” — which is
    essentially the opposite
    of “compound interest” and hence
    not what you mean at all.

    wikipedia on interest

  13. Hi vlorbik, thanks for the comment and feedback. I’d like to understand what you mean about simple interes vs. compound to help make the article more clear.

    For me, I consider continuous interest an extension of simple interest, rather than the opposite. Simple interest has a relatively large term (interest returned over 1 year), while compound interest breaks the duration into many (infinitely small) pieces.

    My meaning was that ln(30) means a period of continuous growth for 1n(10) units of time, followed by a period of continuous growth for ln(3) units of time.

  14. I don’t understand the comment about conversion between simple and compound interest (as described below). If, ln(rt) = ln(2) = .693, then .693…/r should exactly equal the time it takes to double.

    Since .693 is approximated by .72, there is some error in this shortcut, but it is not a function of a conversion from compound to simple interest. Please elaborate on your thinking.

    One caveat: notice how we’re converting between simple and compound interest – won’t this mess up our formula? Yes, it does, but at reasonable interest rates like 5%, 6% or even 15%, there isn’t much difference between simple and compound interest. So the rough formula works, uh, roughly.

  15. Hi, great question. Simple vs. compound interest is pretty tricky and I want to clarify it in a later article.

    There are two sources of error in the rule of 72 equation:

    1) The rounding from .693 to .72
    2) The use of natural log when most interest is actually simple interest. The natural log (ln) assumes continuous growth, but this is not the case for most returns.

    Suppose you have an investment with 5% simple interest.

    To compute your return, you get $100 * (1 + .05) = $105 after the first year.

    To compute your return with compound interest, you’d have $100 * e^(.05) = $105.13 at the end of the year.

    It’s only 13 cents, but it’s a difference that can lead to small errors in how long growth rates will take. (For small rates like 5%, simple vs compound interest isn’t a big deal. For larger rates like 100%, simple interest would be $200, while compound interest would be “e”: $271.281828)

    I do need to take another look at all this and make sure my explanation is correct, it can be confusing :)

  16. This website makes sense of the things that even math teachers have to look up to do correctly.
    Amazing job Kalid, absolutely amazing.

  17. Wow, thanks for the kind words! I’m happy to be able to share my thoughts, I’m glad you enjoyed the site :).

  18. e^3 is 20.08. After 3 units of time, we end up with 20.08 more than we started with.

    Should that read “we end up with 20.08 times what we started with?”

  19. Very informative and helpful. I would appreciate it if you could explain how you would solve this problem.

    Investor A invests $1000 at an interest rate of 5% compounded continuously and Investor B invests $500 at a rate of 8% compounded continuously. If both of them invest at the beginning of 2008, when will the value of their investments be equal ?

  20. @Dave: Nice question. You basically want to solve this equation:

    1000 * e^(.05 * t) = 500 * e^(.08 * t)

    e^(.05*t) is how much growth the $1000 investment has after some amount of time, and similar for e^(.08 * t). If you take the natural log of both sides you can “cancel” the ‘e’ and solve for the amount of time needed.

    @dasickis: Great topic, I’m planning on covering that eventually. First I’ll be writing about what complex numbers mean. I found a good explanation here: http://www.cut-the-knot.org/arithmetic/algebra/Scott.shtml.

  21. You know what? This site was just sooo good that I have to leave another comment. Thank you for this fabulous site.

  22. Cool is the word that comes to mind. Man and all these years I thought that these were some abstract concepts both ln and e. Thanks for letting us appreciate what they really are. I really keep looking forward to your new articles on math .I seriously suggest that you write a book on math and if you dont have the time then take some out of your schedule to write one :)…..

  23. Thanks Mohammad! There’s so much beauty in math, but it gets buried under lifeless proofs. People forget that these ideas emerged naturally, not by someone saying “Today I’ll define e as lim n->inf of (1+1/n)^n”.

    I’ve been thinking about taking the top math/programming posts and putting them into a book… I think it’d be fun. Not quite enough yet, but soon :).

  24. Thanks a lot! Before finding this site, I was trying to understand the meaning of ln and e on Wikipedia, and discovered the same circular logic that you mentioned. Keep up the good work! I look forward to future articles.

  25. Wow – THANK YOU for putting this so simply!!! My friend used this term on the phone earlier. Since I had no clue what it meant, I went to Wikipedia and saw a horribly complicated, technical explanation -but you put it simply!! And true wisdom is being able to communicate a complex topic in simple terms. THANK YOU so much. You made my night!!

  26. Hi Mimi, thanks for the wonderful comment! I’m glad you found it helpful — I agree, there are many subjects that are often explained in an overly complex way. Everything can (and should) be simple :).

  27. Do all bases for logs have to be positive? If not is the log of a negative number possible?
    EG -8 = (-2)^3
    Is this ‘log to base -2, of -8 = 3
    Help please and thanks. JP

  28. Hi Jack, great question. At its heart, a logarithm answers the question

    base^power = desired value

    So I don’t see a reason why you couldn’t use a negative base. The only problem is that certain values are undefined: with regular, positive bases we can’t get to negative values (at least, not without using imaginary numbers).

    With negative bases, certain values are out of range: for example, we can’t get “-4″ since (-2)^2 = 4.

    So, negative bases would work, but would have a different set of undefined values.

  29. Excellent explanation…logical, applicable, and concise. I teach nuclear plant operators a variety of technical subjects/skills (including mathematics), and this is the type of format I strive for in my lessons. Less memorization and more in-depth understanding means more retention! I am going to recommend your site to both my students and my fellow instructors.

  30. Re: comment #41: saying that “certain values are undefined” is IMO somewhat misleading. If you would allow negative bases, such as -2, you would get a very ‘messy’ function.

    If you realise the log is the inverse of an exponential function with the same base, allowing -2log(x) would be the same as allowing -2 as the base in an exponential function: y = (-2)^x.
    Now this last function is only defined for integer x: (-2)^0.5 for example is asking for the square root of -2 which clearly is undefined (assuming non-complex numbers). So the graph of (-2)^x consists of dots which are horizontally spaced 1 apart (alternating above and below the x-axis: the sequence (-2)^1, (-2)^2, (-2)^3 is -2, +4, -8, +16, etc).
    Likewise the graph of log(x), base -2, would consist of dots vertically spaced 1 apart and alternating left and right of the y-axis (-2log(-2) = 1, -2log(4) = 2, -2log(-8)=3, etc) — the reflection of (-2)^x in the line y=x.

    So yes, you’re right that “certain values are out of range”. The problem though is that *almost all* values are out of range, seriously limiting the usefulness of a log function with negative base. There’s a good reason mathematical practice explicitly limits the choice of a base for log and exponential functions to positive numbers (excluding 1, for obvious reasons).

    Your remark that “I don’t see a reason why you couldn’t use a negative base” seems to me to add confusion rather than clear things up.

    Somewhat related observation: when dealing with discrete sequences, all the above is not a problem. You can easily define a sequence x(0) = 1, x(n) = -2*x(n-1), giving the sequence 1, -2, 4, -8, … This demonstrates how such a sequence resembles but is not identical to exponential growth.

    Don’t take my comments to be negative though: I really enjoy your website, including the above article.

  31. Hi Hendrik, thanks for the comment! No worries, lively discussion is one of the fun parts of having a blog. Nobody has all the answers or is immune from mistakes :).

    Yes, looking at it again I should have clarified my response to the negative base. As you say, while it is “possible” to have a negative base, it’s likely not very useful because it only works for a small set of numbers. (As an aside, a negative base may give an interesting graph if you trace out the complex numbers it covers). Appreciate the comment.

  32. Your detailed content really shine. Yours is one of the few that has an in-depth explanation. Maths is, at times, abstract and takes time to figure out the concepts. Here I find it smooth going. Though I have maths blog myself, I like your blog. Good work! Nice learning from you. Natural log and time…. cheers!

  33. your presentation is very good. it has a practical
    insight, but when you come up with 2^n for the doubling equation you seem to pull that out of the hat. for sombody familiar with math its easy to see why but for sombody unfamiliar its mysterious. how did you get there? then you pull out the 100% again how did you get there? then you use the 2^n equation for all other forms. i understand why but i could never derive this and wonder that there is still somthing i dont get.

  34. @Lim Ee Hai: Thanks for the wonderful comment! I’m glad you’re finding the site useful. Yes, math has many beautiful ideas, but they can get hidden behind a wall of abstraction, or lost in a sea of formulas. Glad you’re enjoying it.

    @Mike: Great question, it’s always helpful to see what parts worked and what didn’t. There’s more about the 2^n equation in the article on e, but I’ll try to explain it here.

    Doubling can be considered as growing 100% — if you start the year with 1 item, you finish with 2. So 2 = (1 + 100%).

    If you double 3 times, you could write that as 2 * 2 * 2. That is, after 3 years, you would have 8 items.

    But since exponents are a shortcut for repeated multiplication, we can write 2^3 instead of 2 * 2 * 2. Even better, we can pick any number of years and write

    2^n

    where n is the number of years to use. Remembering that 2 implies a rate of 100% increase, we could replace that with

    (1 + rate)^n

    to get a general equation for any rate and duration. You may also want to try this article on interest rates.

    Hope this helps!

  35. Kalid -

    Great explanation and very helpful. Thanks for taking the time to put it together.

    David

  36. Thank-you! I have a degree in math I have wondered for years why e was used. In all of my classes I was told what e equaled, that ln was the inverse of the e function, and how to compute with e and ln, but never how e was derived! I now have a better understanding and hopefully this will make me a better math teacher in the future! :)

  37. Hi Rhonda, I’m so happy you found the article useful! Yes, there are many concepts that just get thrown out there without the background about *why* it’s useful.

    e/ln aren’t magical, just the result of wondering what would happen if you earned interest as fast as you could (i.e. compounding every fraction of a second). Good luck with your teaching!

  38. Does this make sense? Mike Y said all that about simple vs continuous, and how could it be continuous if it’s simple. But think of the bacteria: If one bacterium can split in one hour, it will have a growth rate of 2, not e, because half of the bacterium cannot start splitting until it is completely split right? Wouldn’t that be the same for populations, so e only applies to things that grow continuously.

  39. I guess I’m wondering, how could a population increase by 1.5, wouldn’t all the bacteria take the entire day to split, or if each were offset, you couldn’t count the incomplete bacteria in the total.

  40. The thing is, if bacteria divided more than once per day, the rate would be greater than 100% per day, but they can’t divide continuously regardless, the time it takes a bacteria to divide would be the time it takes a population to double. I don’t see how this relates to continuous compounding

  41. can someone tel me how to solve the logarithm of a negative number… for example log of -1 to the base 10… i know that the answer is a complex number…

  42. Great question. You can plug in negative logs to google calculator and do ln(-1) / ln (10). So log base 10 of -1 is about 1.36i.

    Euler’s Formula explains how to do negative logs, and I’ll cover this in a future article :).

  43. @Trying to understand: Yep, money and bacteria grow differently (simple vs. compound). However, even bacterial growth can be modeled continuous growth at a different rate. And in reality, there are millions of bacteria at different stages, so the total “clump” looks like one continuously changing amount.

  44. This site is amazing. I have a math degree from a fancy school and I never really understood ln until now. Thanks so much!

  45. Thanks Louis! Yes, sometimes it’s easy to miss the real insights when learning a subject at a deep level. Personally, I’m pretty interested in math and am trying to study analysis, hopefully some “aha” insights will click there :).

  46. cheers man, certinly cleared up some stuff for me. Wish I had this a few years ago at school. I think I get the discrete growth thing, but this really helped me with continuous change. Gonna help me in years to come. You should post on Wikipedia, the maths stuff there is inaccessable from a basic level of knowledge, whereas the article here is practical, much to my liking. Keep it up.

  47. First, let me say I think your explanation of ln is very intuitive and helpful.

    However, your first example threw me off a bit because your terminology is incorrect with regards to simple interest. I see this is discussed a bit in the comments, but the following comment you made is incorrect: “For me, I consider continuous interest an extension of simple interest, rather than the opposite. Simple interest has a relatively large term (interest returned over 1 year), while compound interest breaks the duration into many (infinitely small) pieces.” In the financial world, simple interest means interest on the principal only, with no interest on interest. Compound interest means you get interest on the interest. It can be compounded quarterly, annually, continuously, etc. So constant growth is compound interest, compounded continuously. So, where you say compound interest, I believe you mean continuous compounding. And when you say simple interest, you still mean compounding, but it’s done an an annual or other discrete basis.

    For reference see: http://en.wikipedia.org/wiki/Compound_interest

  48. @G04T_DFA: Thanks for the comment – glad you enjoyed it! Yeah, I make a few typo edits to Wikipedia every once in a while, but generally save the longer stuff for here :).

    @Mike: Thanks for the comment, you’re right: simple interest is like a fixed bond coupon with no reinvestment. As you noticed, I was really talking about the difference between yearly compound interest and continuously compounded interest. I’ll update the article — thanks for the catch!

  49. Thanks for your explanations of compound interest and from whence “e” is derived. Your explanation of compound interest would have saved me from an all night “do-it-by-hand” session 20 years ago. An accountant told me I was correct but it took a very long time to calculate the monies for at least four participants. Henceforth I’m referring a number of readers to you. If there’s anything wrong I’ll have ‘em sue you! Keep up the good work.

  50. @N. Kogneato: I appreciate the comment, glad it was helpful! I might need to add a little disclaimer — “Results may vary” :).

  51. Great write up on a useful concept. I work for an insurance company where we use LN to determine the time at which ‘losses’ or ‘investment’ income can/will grow over time. Our actuaries use various formulas that leverage LN and though I took calc in college, I couldn’t figure out the importance of the number. You’ve done what so many people don’t, shown the relevance of math to real world scenarios. Math, Accounting, and other fields are guilty of using jargon and obscure terms without ever explaining the impact of the terms in ‘real world’ tangible ways. I’ve passed this link on to other co-workers who have all found it useful. Thanks.

  52. @Brett: Thanks for the wonderful comment! Yes, I completely agree — we can end up using an idea for years in finance or engineering without understanding what it really means. It just becomes a magic equation we throw numbers into :).

    Glad you and your co-workers enjoyed it!

  53. Great explanation for ln, but you already knew that…

    I’m working on a problem in my quantum mechanics class, the Baker-Hausdorff lemma, and I can’t figure out how they’re expanding the series! Could you give me a point in the right direction?

    z = ln [ (e^A)*(e^B ] = A + B + (1/2)(AB – BA) + (1/12)(A^2B + AB^2 – 2ABA + B^2A + BA^2 – 2BAB)

    Is this some sort of twisted Taylor Series expansion?

  54. @BatManda: Thanks — great question. I don’t know much about that formula unfortunately, it does look like a crazy taylor series. It seems strange to break the equation apart when you could simplify it like this:

    =ln[ e^(A+B) ]
    = A + B

    But clearly I must be missing something :). You might try the math/physics forums (there’s a section for quantum): http://www.physicsforums.com/

  55. First, let me say that I found your articles on e and ln to be very enlightening. Thanks for taking the time.

    That being said, I agree with the general consensus that your explanation of 100% continuous growth is confusing. I’m not sure if this is valid for all uses of e, but I prefer to think of all cases of continuous growth to be 100% while only time varies. In this way, all talk of percentages can be left behind (as I’m not sure the concept of fractional continuous growth really applies to real world situations) and allows us to simply say “continuous growth.”

    Also, I am surprised that no one mentioned the case of continuous growth that everyone who’s taken high school chemistry should know about: dG = -RT ln K which relates the reaction rate constant K to Gibbs Free Energy. Continuous growth is essential to any topic relating to chemistry including phase change, thermodynamics, heat and fluid transport phenomena, and a wide range of other topics. Math is a tool of science, not the other way around.

  56. what about the power property of natural logs.. how is this derived?

    ln(b^t) = t ln(b)

    this is a bit confusing to me

  57. @Luke: Thanks for the comments! I think I need to revisit the continuous growth explanation as it can be really confusing. I see continuous growth of r% meaning:

    * 1 unit will product r% interest after 1 time period
    * That r% interest, as soon as it’s made, will produce r% on itself after 1 time period

    So, the “money that money earns, earns money”.

    The problem is the difference between APR (annual percentage rate — your “growth rate”) and APY (annual percentage yield — the total amount you change as a result of the compounded interest).

    Because it’s e^r*t, you could see it as 100% growth letting time vary (as you mentioned), or 1 year letting the growth rate vary. But still, I need to think about ways to clarify.

    Thanks for the relation to chemistry, it’s been a while since I’ve looked at that and it’s nice to have reminders :).

    @Anonymous: Great question. ln(b^t) means: How long does it take to grow to b^t?

    Well, growing to b takes ln(b) time (for example, growing to 3 takes ln(3) time). Growing to b*b takes ln(b) + ln(b) time [grow to b, then grow to b again -- to grow to 9, you grow by 3, then grow by 3 again].

    So, growing to b^t means you just add up the time to grow to b [ln(b)] “t” number of times: t * ln(b). Hope this helps!

  58. I’m taking a financial structuring course at an undergraduate level… always knew how to use e but never really grasped what the hell it meant! thanks a lot for this it really helped refresh and consolidate my knowledge of this function!

    thanks again!

  59. @RP: You’re welcome, I’m glad you enjoyed it! The true meaning of e didn’t hit me until much longer after I “learned” it also :).

  60. Many research papers use the natural log for items such as total assets or sales revenue as a size variable. Why is the natural log a better size measure versus just using the actual dollar amounts for total assets?

    Phil

  61. Hi Phil, good question. I’d be surprised to see natural log taken on a straight number (total assets or sales) — usually it’d be used for a growth factor or a series. It could be used to rescale numbers to a “human” factor (like a Richter scale or Decibel lets us talk about large variations in a smaller range). But there may be another interpretation of the natural log which would make sense for the raw number.

  62. U’m In Pre-Calc and sadly when the teacher talks thats all she does I get better explinations from other students, and this was a new topic, which of couse i thaught i would never grasp thank you for this…:(…i sorta wish you could teach me now…but i digress and return to theact of thanking you and hope that you cover some other topics i’ll have to come.

  63. I just want to THANK YOU because your article was so just so much more straight forward then my text book. This article explained the concepts and a more comprehensive way then I think even my teacher could explain it. THANK YOU AGAIN and keep wrtiting!

  64. @Kid Crew: Glad you enjoyed it! Hopefully I’ll cover the topics you’re learning about. Really happy to hear things are clicking.

    @Calculous Crushed: You’re welcome! I had a similar reaction when learning about logs and exponents, and only “got” them much, much later after learning about them. Thanks again for the encouragement.

  65. I dont quite get the the rule of 72 thing could explain that to me. I dont get the idea of having rate and time always getting the same answer.
    Thank you. (great article, btw)

  66. Can you please explain how log returns work (ie IBM starts the day at $5 and finishes the day at $6…the return would equal ln(6/5))…i believe this type of calculation is used in finance to calculate returns. This is different than (6-5)/5….I understand the log is using continous growth, but can you explain how to think about it, and also why most firms use one technique rather than the other.

  67. Hi Jeff, interesting comment! I’m not certain of the reason, but here’s what comes to mind.

    e and natural log let us discuss growth in a way that is “flexible”. Using your example, (6-5)/5 represents 20% growth in a day. ln(6/5) ~ 18%, which is the equivalent continuous growth rate (which is less than 20% because continuous growth earns “interest on interest” and can catch up).

    With the continuous growth rate, we can say that having 18% growth for 1 day will have the same impact as 1% growth for 18 days. Or 2% for 9 days, or 3% for 6 days, etc.

    If we use 20% standard growth, we can’t easily swap out the numbers since rate and time can’t be mixed. Natural log lets us convert rates to “standard” format which makes it easier to do mental “what-if” scenarios. That’s one application I can think of :).

  68. Hi Kalid, brilliant explanation, really enjoyed it.

    I was thinking about cell division.

    If I understand correctly, we cannot use our formulas if there are less than 100 cell divisions over the time period under consideration; if there are less than 100 cell divisions the formulas will be too inaccurate.

    See Kalid’s explanation here

    If the cell could divide in zero seconds then our formulas using e would be 100% accurate.

    A cell division every zero seconds corresponds to a growth rate of infinity%

    Using our formula above The time for the cells to double would be given by
    time x infinity = ln(2)

    time = zero seconds.

    This is a trivial result, but ironically this is the only time the formulas are 100% accurate.

    The rest of the time, when we are dealing with real life systems, the formulas are so close to accurate that errors can be ignored.

    Have I got it or am I way off the mark?

  69. Yes yo, bn a great pleasure reading the article. It opened me up a bit, thanx to you proficiency.I wish to kindly ask if you can be able to explain to me how to hanndle exprassions of the form :Ln(x+y). Just how this can be simplified will do me good.
    Thanx in advance!

  70. @chris: Great question! Here’s the key:

    Real-life systems can’t grow perfectly continuously. But, they can be approximated by some continuous function.

    An analogy may be how we try to make wheels on a car “round” — that is, follow the equation of a circle (x2 + y2 = r2). Of course, at a molecular level the atoms aren’t exactly on that line, but the above equation describes the shape of the wheel very well.

    For cell divisions, individual cells may not divide at a continuous rate (or even at 100 times per second). But the whole “lump” of cells may appear to grow at some rate — we have a lumpy goop that is growing over time.

    e^x can then approximate the “expected” growth rate of that goop, like the circle can model the “expected” location of atoms in the wheel. So, you don’t need your cells to grow at any certain rate, you can find some rate that approximates the way it grows. You can find some growth rate x to make e^x fit the curve of your actual growth. If your goop grows 2x after a unit of time, your equivalent continuous rate is ln(2) = .69, and can be modeled by e^.69x. Hope this helps!

    @Sehluko: Hi, I’m not sure there’s an easy way to simplify ln(a+b) by itself. If there are other terms ln(a+b) + ln(c+d) you can do = ln([a+b]*[c+d]).

  71. Kalid,

    Great work.

    Question: as you know, we think of risk and return in finance. Let’s assume we have an expected rate of return (on which we can apply e assuming the rate is constant or close to constant). But let’s further assume that returns are uncertain and normally distributed around the mean (with a known variance). As my variance in returns increases, what happens to e?

    Thanks!

    David

  72. Thanks for your article. It helped a tremendous amount. My unit guide describe all of ln and e within one side of a standard a4 piece of paper. It was incredibly useful! My unit jumped from ln and e into exponential and logarithmic models (Ln Reg). Thank you very much, and I’d be happy to read more!

  73. Hi Kalid
    thanks so much for your response. What I was trying to say is elaborated below:

    Let us say we have a sum in the bank invested at 100% growth per year compounded every 6 months. If we dive straight in with our formulas involving e we would say:

    We have 100% growth over 1 year rate x time = 1
    Therefore growth factor = e^1 = 2.71828182846

    However using our compound growth formula:

    Growth factor = (1 +1/2)^2 = 2.25

    A fair size inaccuracy has crept in:

    So we always have to be mindful of how many times the we have compounded over the period in question.

    If there was continuous growth, of course, there would be no problem.

    Now let us consider rates different from 100%; for example, a growth rate of 1% compounded 100 times over a year.

    We can jump in with our e formula and say
    that the growth factor will be:
    e^(1/100) = 1.01005016708

    However, if we use our compound growth formula we get a growth factor of:

    (1+.01/100)^100 = 1.01004966209

    Why the discrepancy?

    Well let us recall how we derived the e formula for growth rates that are different from 100%

    We said that a continuous compound could be approximated to a number of discrete compounds.

    Applying this to the problem above: we will approximate 100 discrete compounds to 1 discrete compound.

    In other words we are stating that:

    1+.01/100)1^100 is close to

    (1 + .01/1)^1

    We then rearrange this to give:

    (1 + 1/100)^(100x(1/100))

    But (1+1/100)^100 is nearly e

    so the answer must be e^(1/100)

    right so let us examine the areas where we have allowed inaccuracies to creep in.

    first we said:

    1+.01/100)1^100 is close to

    (1 + .01/1)^1

    In fact it is a little bit larger

    Then we said:
    (1+1/100)100 is close to

    e,

    In fact it is a little smaller

    So we’ve exaggerated in opposite directions allowing us to get an answer but it is an approximate one.

    If we had continuous growth instead of our 100 discrete compounds over the time period in question, all of these inaccuracies would disappear.

    We can see this if we look at the same problem again but with continuous growth over the year.

    Let us approximate an infinite number of discrete compounds to 1000 compounds

    The compound growth formula gives:

    1+.01/infinity)^infinity

    This is close to

    (1 + .01/1000)^1000 (closer than the last approximation because the series converges.)

    This can be rewritten as:

    (1 + 1/10000)^(100000x(1/100))

    making our e formula even more accurate because (1+1/10000)^10000 is closer to e than (1+1/100)^100

    You can use your imagination to determine that as we approach continuous growth the e formulas become more and more accurate.

    However very few systems in the real world exhibit continuous growth so caution must be exercised.

    I am an amateur mathematician so this could be a load of rubbish please chip in with your two cents. Thanks for such a brilliant web site Kalid.

  74. FWIW, in my class, the logarithm was defined as the antiderivative of 1/t, and the exponential map as the inverse of the logarithm.

  75. Now the integral of 1/t = ln(t)

    From Kalid’s explanation above that means that the (RATE x TIME) for something to grow, if it experiences continuously compounded growth, is given by the area under the 1/t graph from t = 1 to t = final size.

    This can easily be shown mathematically but I was wondering if anyone has got a ‘better explained’ intuitive reason why the (RATE x TIME) for exponential growth is equal to the area under the 1/t graph.

  76. Good job with the explanations. I do think a visual aid would be fantastic. A little graph. Maybe plotting y = ln x and y = e^x. Relating your explanations to a visual would certainly help me. I have had to read over and over some of your explanations (needing more clarity before moving on).
    If you have time that is. Inbetween explaining other things.
    Thank-you

  77. Further; You have not linked the concept of ln to e very well in terms of time to reach a certain level of growth. Consider the following

    At 100% (continuous) growth, time to double can by definition of e: 2x = x.e^t, where x is the initial amount.

    Lets get rid of x (/x to both sides): 2 = e^t
    Using laws of logs: ln 2 = t ln e^1
    ln e = 1, this leaves ln 2 = t

    For those of ous who have a good understanding of ‘e’ gained from your last tutorial, this helps to tie it together. Hope you agree and consider it.
    Thanks again

  78. assuming 100% growth rate, simplified statement?

    e^(time) = amount
    ln(amount) = time

    explanation:
    e^(time*100% growth) = amount (at 100% growth)

  79. I am building an mapping application where I need to convert geolocation data (longitude / latitude) to x/y coordinates and vice-versa. I am having trouble understanding how I can solve for latitude given y using the Miller cylindrical projection method. I tried putting an image embed code below but if it doesn’t show up you can see the equation here: http://en.wikipedia.org/wiki/Miller_cylindrical_projection

    I am interested in learning how to solve for this but the log math is still a bit abstract for me. Can anyone help clarify this problem?

  80. Wow, I never liked math in school because I didn’t understand WHY the things like “e” and “ln” were practical. I guess I had a hard time memorizing mind-numbing formulas with teachers whom, I am guessing, did not even understand it quite so intuitively themselves. Thanks for sparking my interest in something I did not think I could effectively utilize in life.

  81. @phil: Awesome! Yes, there’s so many things we learn by rote which can just kill our interest in a subject, glad you’re enjoying it again.

  82. Thanks for such a wonderful explanation of the exponential function and its inverse. I never understood how my calculus books came up with this number. Because they managed to avoid the “why’s” with rather unnecessary calculations and problems.

    Thanks again for such a simple and intuitive explanation. This should be published!

  83. @awesome: Thanks for the wonderful comment! I’m happy it made sense to you after all this time, it was such a relief for me too when it finally clicked. And thanks, I hope to publish this one day :).

  84. I tutor college students. This is exactly the sort of explanation they need and I strive to provide. No, it’s not mathy/technical, thus it’s open to attacks from the “what ifs” and “but you fail to recognize” crowd. But this is the kind of explanation that will help the average student truly understand the concepts and–much more importantly–retain the information. Good job! I might even steal a couple of ideas from you. I hope you don’t mind.

  85. @Mike: Thanks for the note, and you’re more than welcome to use the ideas here in your tutoring! I want to help as many people as I can to have that “aha” moment — glad you’re helping it along :).

  86. I worship you!!! 12 years of grad school and 2 years of undergrad have not come close to what you have done here to help me understand

  87. @Anonymous: I know what you mean! It took me a long, long time to really understand e / ln to a level that made intuitive sense!

  88. Man, I do not even know where to begin. I love your site and articles. e and ln articles are incredible, real art. In 5 min you have answered some questions that bugged me for years. You have amazing talent in teaching and I would love to see math/calculus book written by you some day. Engineers, computer science and math enthusiasts do not need “rigor”, they need real knowledge and understanding of the subject. And honestly I have never read anything even closely as good as what you have put on this site. Please keep it up!!!

  89. You didn’t mention that ln(0) is also undefined, not just ln(negative number).

    But great site! Now I can go teach my teacher!

  90. Wonderful article.

    In response to comments 48 and 49: This is a little technical, but it’s a really neat point, and maybe (just maybe) it’ll get some more people interested in math. If you allow yourself to wade into the world of complex numbers, taking the logarithm with a negative base yields a very interesting result. As you said, log(base -2) only maps onto the integers for certain cases; for example -2^0=1, -2^1=-2, -2^2=4, etc. But for -2^0.5, you get i*sqrt(2), a purely imaginary number. Continuing, -2^1.5=(-1)^(1.5)*2^(1.5)=-i*sqrt(8). Now, if we plot these in the complex plane, we have a point on the real axis at 1, a point on the imaginary axis at 1.414, a point on the real axis at -2, a point on the imaginary axis at -2.828, and a point on the real axis at 4. The kicker now is to figure out how to handle a number like -2^(1/3). This equation basically asks you to look for a number that, when cubed, gives you -2. You can check that 1/2 + i*sqrt(3)/2 does just this. This gives you a new point on the graph at 0.5 on the real axis and 0.866 on the imaginary axis. If you haven’t tuned out yet, you’re probably starting to see a pattern here. In fact, if you keep graphing points like this, you get a figure known as a logarithmic spiral. This type of spiral is seen all over the place in nature. Wikipedia gives great pictures of the spiral arms of a galaxy, the inspiral of a hurricane, and a nautilus shell to name a few.

  91. Wow, im a student at syracuse univeristy and I have the worlds worst precalc prof. and thoroughly dimishing my interest in this class, however, I cant give up cuz Im spending 50,000 to go here and plus im premed. but let me just say if my prof could get up off his high horse and take the stick out of his rear behind and teach logs like this or anthing like this, I would actually be learning something and wouldnt have to resort to going online, when Im paying this much. thank you sir/ma’am. this has helped enormously! :)

  92. Hello, I would like to say that this post has helped me enormously,
    I go to Syracuse University and I have the worlds worst precalc teacher. If he could get up of his high horse and take that stick out of his rear behind, then he could actually succeed at being a teacher. Why dont people teach you the basics, stop with the formulas and TEACH people. well thanks so much sir/ ma’am. :)

  93. WOW! Thank you Kalid!
    I’m a university student in New Zealand, and I only took maths for the first three years of high school. Now that I’m back studying I’ve realised that calculus wasn’t just a series of classes that I bunked in high school maths, but a subject that I might (shock, horror) actually need it for my Economics degree.

    I stumbled across your articles while I was quite literally in tears about the idea of learning calculus – I love economics but I was even considering changing my major because it just seemed too hard, and my head hurts the moment I look at a formula. I’ve been reading various stuff on your site for a solid hour and my head doesn’t hurt a bit! I intend to send a link to my lecturer in the hope that he will recommend this site to all the other social science (rather than commerce) students who are struggling to get their heads around the maths of economics,the most beautiful subject in the world.

  94. @Izzy: Awesome! Really, really glad it helped! I completely know what you mean about the feeling of frustration — it seems math evokes it more often than other subjects :). I’m thrilled you stuck with it and found value in the articles here — thanks for sharing it!

  95. How do you estimate “negative feedback” like the slow down of growth in bacteria when they run out of media. Or when the growth of a real estate development runs out of attractive sites. My guess is you need some parallel process and use empirical corrections.

  96. ” * e^x is the amount of continuous growth after a certain amount of time.
    * Natural Log (ln) is the amount of time needed to reach a certain level of continuous growth”

    Aren’t you scrambling up two different concepts here? e^x is not the amount of “continuous growth” — it is the AMOUNT, period. x = rate * time. One of the first things I learned in elementary algebra is that rate * time = distance (“amount” is just another term for “distance”). e^x is the AMOUNT accrued “after a certain amount of time”.

    Conversely, ln is not “the amount of time needed to reach a certain level of continuous growth” — it is the amount of time needed to reach a certain amount.

  97. @Robert: Great question, I need to think about how to clarify this. If someone says “I’m growing at 100%, how much will I have after 3 time periods?”, there’s a few answers:

    * e^3 if you are growing continuously
    * 2^3 if you compound every time period
    * 1 + 1 + 1 + 1 if you assume simple interest (no compounding)

    While all answers are some type of “amount”, I find it helpful to qualify how that answer was determined (did we assume continuous, compound, or simple interest?). There may be a better way to phrase it, but the key is to understand how the amount was computed.

  98. Hi. I will read more of this article today (just been reading your article on e – it’s truly excellent). I see someone else has mentioned the Gibbs free energy equation here. Is there any chance you could explain why the natural logarithm is used there??? The “RTlnK” bit. The natural log of the concentration of products over the reactants ln[products]/[reactants]? I don’t understand why that is done. Why, why, why? Someone please help :( I will worship you forever.

  99. “If the population of Wyoming is changing exponentially and in 1980 the population of Wyoming was 469,557 and in 1990 it was 453,588 and in 2000 the population was 493,782. What function P(t)= P*e^kt is suggested for the given data for 1980 and 1990?”
    Can you please show me how to answer this problem it is really confusing me, because the rate seems to change from negative to positive.

  100. When I first read this article, it seemed to make sense. But then I thought I would just go over it again and now I see something that doesn’t make sense to me (perhaps because I’ve taken it out of context). You wrote:
    “100% return for 3.4 years is 30x growth.”

    Say each year on New Year’s Day, I place a bet at 1:1 odds, which means I earn 100% return on my money. If I start with $1 in year one, then on Jan. 1 of year two I now have $2. On Jan 1 of year three I have $4, and on Jan 1 of year four I have $8. That’s not 30 times growth.

    If a rate of return is defined as 100% per period, then it’s 100%. It’s not 100% compounded continuously. 100% per year, compounded continuously, is 271.8% (or so) per year.

    So intuitively, I’m having trouble arbitrarily converting growth rates to some power of e that in effect, change the rules in the middle of the game. It’s more intuitive to me to just say that e to the 3.4 power is 30, and that ln 30 is 3.4. I really do appreciate the idea of intuitive and imaginative explanations, and I actually really dig your site and think your the entire concept is really bold and timely. This article just didn’t work for me, though.

    Logarithms are just high powered “demagnifiers” that let us work with really large numbers over time on a human scale. The natural log of a number is just that number’s value as a power of e, and that works for certain types of models and not others. It gives you the time needed to reach a certain level of growth but ONLY if that growth is modeled by e, or it could be used to find the growth rate as a power of e given a particular time frame. I hope I got that right! ;-)

    So e is an interesting number but so is 360, right? We could just define e as 10^.4342944819 and call it “abracadabra” and maybe that’s just what’s needed to demythologize it and make working with it more intuitive.

  101. @Fred: Great questions!

    For the doubling example, it’s important to recognize there are two rates: the “input” rate and the final “output” rate. In the betting case you describe, we agree to only look at the total output (doubling) at perfect year-long intervals. For most man-made things, this is simple and works.

    For most natural processes, however, we only know the current/instantaneous rate: i.e., I’m constantly growing at a rate of 50%, where does that put me in a year? It won’t be 50% higher because in 1 minute I’ll have grown a bit and will then be growing at 50% of that *new* rate. e lets us work out what the final output will be.

    Unfortunately, “interest rate” is an overloaded term. In the banking industry, they talk about APR (annual percentage rate) and APY (annual percentage yield) – APR is the input, APY is the output (what you actually pay). Simple, compound, or continuous growth patterns can take the same APR (of 10%, say) and return a different APY (total result after the time period).

    Yes, I like the note about “demagnifiers” — the natural log lets us “undo” the effect of exponential growth. But here’s the tricky thing: any amount of growth can be represented by e! Even processes that didn’t start growing continuously (like your doubling) can be _described_ by an equivalent continuous process. That is, even though 3 * 4 = 12 isn’t a square number, 12 can still be seen as sqrt(12) * sqrt(12). The neat thing about continuous growth/ln/e is that it’s a standard reference point to compare different growth rates.

    e is interesting because it’s the basis for this standard reference point. Indeed, we could say e = 2.71828… and memorize it (which is often done!) but it’s more illuminating to say “e is the number when you perfectly compound 100% growth for 1 time period”, just like “pi is the circumference of a circle of diameter 1″. Just giving the raw result can hide the reasoning behind why the number is so useful.

    Hope this helps!

  102. Thanks Kalid. BTW, in my example I miscalculated the effect of continuous compounding as “271.828% growth.” I neglected to subtract the principal when calculating the percentage increase. A stated rate of 100% return over a period, compounded continuously actually represents 171.828% effective growth at the end of that period. (Oops.)

    The contradiction of course is that if the stated rate has a major asterisk, “compounded continuously”, then why even bother quoting the stated rate? If I borrow at 10% per year, there had better not be any fine print that says “compounded continuously.” That’s why banks are required to state the APR on a loan and not just the interest rate. Ignoring fees, the real interest rate paid is determined based on a combination of the stated (simple) rate and the method of compounding.

    Of course I was downplaying the importance of e. e^x is the only function that equals its own derivative, and it’s essential for computing probabilities. It’s spooky the way it keeps popping up in all sorts of real-world situations, and probably will for a very long time.

    I like the idea you presented in a prior article that “e represents the idea that all continually growing systems are scaled versions of a common rate.” I also like the way you point out here that the natural log’s value is composed of the factors of rate and time, a very helpful way of thinking about continuous money growth.

  103. Also: I do see what you’re saying about the natural log being a useful benchmark for comparing different growth rates, whether continuous or not.

  104. You, Sir, are on to something.

    The authority of insanity has held sway over mathematics for far too long… Those of us who in our hearts know things should be understood are generally told we are wasting our time, worrying our poor little heads, or even worse just plain stupid because we are not repeating parrots. The message given is to put up, shut up, give up, accept without justification, blindly do whatever everyone else is doing. Accept their underlying premise that life is basically miserable. Bit hyperbolic here, but I’m fairly sure this is the same process that leads to moral atrocities of every kind.

    Thank you for your intellectual courage and integrity, long may you continue your excellent and inspiring work.

  105. @chelsey: Thanks — I wish the emphasis on school was on really “getting it” also.

  106. I went in search of an explanation to ln and e since the textbooks and teachers both use the same non-explanatory replies, but you…wow…absolutely wanting people to understand and help them. Thank you so much for simplifying ln and e as well as the rule of 72. I have always felt teaching the principle is key to fully grasping the whole concept and people will be smarter for really knowing what this means and how to use it easily 10…20…30…years from now. That is true teaching. Beautiful. THANK YOU! (not shouting only expressing my appreciation)

  107. I read this piece and the one on “e” a while ago, and am re-reading it (with my copy of the ebook, which everyone should get). Please keep churning out the articles. A request related to this topic: how about helping to explain the intuition behind using logs and other ways to re-express data to get a linear relationship, I have a tough time wrapping my brain around when it’s “OK” to transform data like that. Other topics in statistics would be great too.

  108. Here’s a brain teaser. When you punch ln(i) into your calculator, you get pi. But e raised to the pi power is a real number (=23.1406926). So ln(23.1406926) = pi. Could it be that ln(i) actually has many solutions?

  109. @AC: Thanks for the encouragement, I’ll keep the articles coming as the insights happen :). I’d really like to dive into e and ln some more, especially with their more practical uses for log-log plots, etc. I still don’t have a good intuitive feel for that yet, and it’s something I need to look into — thanks for the suggestion.

  110. @Snorkle: Actually, ln(i) should really be “pi/2 * i” — you can go 1/4 of the way around a circle to end up at i. It might be a mistake on that calculator ;).

  111. I would have kissed you a hundred times for such a nice article on the topic, if you were near me.

    ~*~*~ Bundle of Thanks ~*~*~

  112. Thanks a million Kalid for the great explanation…the best I’ve come across on the topic….thanks again…

  113. This site is incredible. Leagues beyond anything my mediocre high school mathematics instructor was ever able to put together.

  114. how would i work out
    (0.065t)
    500e =1000

    i know i have to put ln on either side but i dont know what to do with the 500.

  115. I want to echo what Admirer said. I know how much time it takes to write and maintain a technical blog. You aren’t making money from it. But, if I had my way, I’d at least be able to buy you a cup of coffee.

    You will never know the good you’ve done.
    thanks

  116. @antimom: Thanks for the encouragement! With some luck, I may be able to turn this into a full time gig someday :).

  117. * ln(negative number) = undefined

    Undefined just means “there is no amount of time you can wait” to get a negative amount.

    Good article, but this is wrong. The answer would be an imaginary number, not undefined. Try at wolframalpha.com if you don’t believe me.

  118. I pulled your web site for help and have found anything but!!!!! The Rule of 72 – ???!!!???!!!??? What are you saying? That you round up to the next number that has several divisors? You could not have been any more unclear if you planned to be so.

  119. Nice article, thanks a lot also for the one on “e” just very well explained.

    but would be great if you could also explain why you would sometimes calculate ln of returns e.g. stocks ln(stock price 1 / stock price 2). My book says something about a brownian motion and to standardise it blabla. I am sure you could explain it much better in an article on returns and volatility

    Thanks again!

  120. Thanks, I have spent my adult life not understanding this. I think I slept through that class in High School and was able to fake it in college. I feel a strange sense of peace and satisfaction.

  121. @return: Thank you — that’d be a good article. The standardization comes in because e is a universal way to think about growth (how fast am I growing this instant?) instead of having to pick an artificial measuring point (monthly, daily, yearly growth measurements, etc.).

    @Dean: Thanks! Yes, there’s definitely something very satisfying about finally understanding these ideas which have eluded us.

  122. How do I calculate my time to reach say 25000 dollars when my starting amount is say 2000 at 3% annual interest? This isn’t covered very clearly in your otherwise excellent presentation.

  123. Does anyone know how Euler made this formula for ln x? :

    ln x = lim n–>inf n(x^(1/n) – 1)

    It give accurate results, but I havent seen it used much and it’s probably just as important to the similar type of formula for e^x and knowing how it was derived.

  124. (continued)

    The analysis in the article is nice. I might have something to add to these ideas.

    N = ce^x where c will undergo continuouly compounded
    growth for x periods of time. But if you want to
    find how many periods of time it will take c to
    grow to N then you must first divide both sides
    of the equation by c:

    N/c = e^x

    Ex. N = (5)e^2 = 36.945 : x = 2

    Now consider ln 36.945 = 3.609

    If you divide N by c first:

    36.945 / 5 = 7.345

    ln 7.345 = 2

  125. wow great way of describing ln and e. I have sat through many lectures with many different profs to explain those two constants and never have I seen it explained so perfectly. Great job!!

  126. great article, kudos on the simple language used, just read the “e” page and all this makes sense now,

  127. Kalid: Thanks for your great efforts to make math more intuitively understandable.
    But in this article as well as the one on e^x, you insist on calling e the “growth” when it is clearly the amount of growth +1. Or growth = e – 1.
    That you know this perfectly well you demonstrate in a couple of places in the e^x article:

    “(Aside: Be careful about separating the increase from the final result. 1 becoming e (2.718…) is an increase (growth rate) of 171.8%. e, by itself, is the final result you observe after all growth is taken into account (original + increase)).

    We get a formula by using 3 periods in our growth equation:

    \displaystyle{growth = (1 + 100\%/3)^3 = 2.37037…}

    We earned $1.37, even better than the $1.25 we got last time!

    So why do you then turn around and call e the “growth” in other places in your article? This becomes even more confusing when one tries to understand the calculus concept that the “growth” in the e function is always equal to the slope.

    Ron N.

  128. You have a gift, man. Natural log finally makes sense! If only I had found your website / ebook when I was going through high school and college (though you probably didn’t have it all written out between 2000 and 2006).

  129. I found this article quite helpful, but one point left me a little fuzzy until I realized (or think I realized) that rate*time=ln(growth) is what you were actually doing when you said rate * time = 3.4. Also, it would have been much more natural to rearrange the numbers or words in the formula so they were in the same order.

    In other words, you said rate * time = 3.4 and then said 3.4 years * 100% = 3.4; however that used the order time * rate…

    Thanks for making this easy to understand overall, though. This was the first time I ever attempted to learn about LOG, so the fact that I think I understand it already is pretty good. I am just trying to find a way to fully apply it to my current life. More examples of applications might be useful, but not necessary.

    :)

  130. One of the best site I ever found in explaining mathematics in its simplest way.. Thanks khalid

    Keep up the good work and may your tribe increase..

  131. @Hash: Thanks for the note! Yep, I’d like to do more articles for different age ranges, and maybe segmented by category.

  132. @Ron: Thanks for the comment — I think I need to go through and do a cleanup pass to normalize all the language used :).

    @Dustin: Ah, great feedback! Yes, I should have it match up rate * time, 100% * 3.4 years would be a lot more clear. I agree more examples might be helpful, I might do a follow up about using e / natural log in the real world, and how to get more comfortable with it.

    @Evolution: Thanks!

  133. I think the section on Using Natural Logs with Any Rate would be a bit clearer if you worked out the example:

    We want 30x growth, have 5% rate , how long with that take?

    ln(30) = 3.4 [3.4 represents rate* time]
    3.4 = 0.05 * time [plug in rate, represented as a decimal] [a rate of 100% = 1]
    3.4 / 0.05 = time [divide both sides by 0.05]
    68 = time

  134. Very nice article and website in general. You and I Kalid think alike regarding math :)

    I was thinking about two things after reading the article:

    1. I’m kind of missing a justification that “at reasonable interest rates like 5%, 6% or even 15%, there isn’t much difference between yearly compounded and fully continuous interest”

    2. You may have mentioned this in another article in calculus (or not), but I think the fact that d/dx[e^x] = e^x and perhaps even d/dx[e^(kx)] = k*e^x would be an interesting and relevant point to derive intuitively in either this article or the one concerning e.

    But as I said: good job! ;)

  135. @Gandalf: Thanks for the feedback!

    1) Yes, I wasn’t very specific about the low interest rates. The intuitive explanation is that at low rates like 5% or even 15%, you aren’t earning enough “interest on interest” for it to make a big difference in the calculation. For example, e^(.05) = .05127… which is vey close to .05, the amount without compounding.

    2) d/dx e^x = e^x is a great way to see e — check out this article for a few more examples on it: http://betterexplained.com/articles/developing-your-intuition-for-math/

    THanks for the suggestions!

  136. @Gandalf: Great idea, I’d like to cover that in a future article! You can see the hint of the e expansion in there.

  137. This explanation puts everything I learned on Khanacademy to shame. Keep up the good work, and please continue to post on more math topics. I would love an intuitive explation on the quadratic formula, which I will be using a lot in chem 2.

  138. Kalid.. U r awesome with these explanations, we’ve been stuck our whole childhood that what is this means, why this means this, but u just crystal clear all the mist or myths that we’ve been through. We just keep reciting that say “Pi is 3.14 or e is 2.714″ but u made them crystal clear in our mind that we would never forget… we in real world just stuck between simple mist in our everyday life of maths that why this meant that or what is that meant this.. but u just clarify all this….. 5 Thumbs up and keep posting articles about the regular maths problems like this that leads us nowhere but to keep reciting the value or formula.. so please keep submitting articles on our daily confused problems of math… Thanks again…

  139. @Aadil: Thanks — I really, really want everyone to be able to understand math at an intuitive (not memorized) level. Appreciate the comment!

  140. hey..Kalid thanks a lot for this awesome work please let us intuitively teach how to calculate the values of logarithms and natural logarithms specially natural logarithms as soon as possible…..Thanks in advance waiting for ur reply……

  141. Thank you thank you! Very well explained!! And being so patient in explaining further in the comments portion. Kudos to you! :)

  142. @bc: Awesome, thank you! I love the discussions, it helps me think about the topics in new ways. Having discussions is a fun way to cement our understanding :).

  143. @Aadil: Thanks for the comment — I think the easiest way to calculate logs is to use a calculator :). But, there are probably some methods by-hand that I should learn. Might make for an interesting post if there’s some intuitions to be found there!

  144. Hi,

    Can someone help me with this?

    1369*e-ln2/6.73 * time

    i would appreciate a step by step or if someone can write an excel formula for that. please.

  145. Hi people,
    I’m having a bit of an issue with what ln does when you apply it to certain variables.
    Writing a paper at the moment and the professor suggested that I used the natural logarithm of the company size instead of the size itself in a regression.

    My main issue is: What is the change in definition when you use the LN on a specific variable?

  146. Dude, I have a master’s in chemistry and now teach it but have never had this kickass of an explanation of e, nor this clear of an understanding. thank you, I can now pass on a bit of this understanding to my students

  147. @mike: Awesome! One of my goals for the site was to give teachers analogies they could use, really happy you’re able to share it!

  148. It makes sense that the exponential function to base e is the solution to the differential equation dy/dt = ky whereby the rate of change of y is proportional to the value of y at any instant. This is because the solution to this differential equation must be a function whose value is continuously evaluated and compounded; every zillionth of a second a set percentage of the value of y(t) is added on. Thus it makes intuitive sence that ln(y) is the integral of 1/y and hence the solution to dy/dt = ky is y(t) = Ce^kt.

    What I find mindblowing is how the number e so easily springs out of algebraically manipulating the logarithmic function!

  149. @Cameron: Neat way to look at it — yep, the differential equation says “instant by instant, this is how much I’m changing” and e^x accounts for continuous, instantaneous change. I’m actually not sure how you can get e from algebraically (vs. calculus…ly) manipulating the log function, but I’m curious — feel free to share!

  150. @Kalid: Yeah I meant the number e springs out of differentiating the log function from first principles, which only requires basic algebra.

    Incidentally, do you happen to know who first discovered ln(x) as the integral of 1/x? I’ve been searching for the answer for ages, but haven’t found anything.

  151. @Cameron: Thanks for clarifying! Hrm, offhand I don’t… I’m just going to guess Euler :).

  152. “ln(negative number) = undefined” seems confusing to me as it implies that x could be 0, and ln is defined only at x > 0. just my thoughts

  153. Very nice. I was looking for a way to describe \displaystyle{ln(x)} and \displaystyle{e} to undergrads for next fall and I think I may have to refer them to this site.

  154. Although it may not be derived intuitively, the old formula “The log of the products is the sum of the logs” is poetic enough to stick in the memory like a good mnemonic.

  155. Excellent! Explaining the relation of math basics to real life has been the weakest point in all math books I’ve ever opened. Surprising, too, cause it just happens to be The Key for us “non-geniuses” to really understand math. Why, actually every such point was originally understood in relation to some real life problem! Yet quite often it happens, that although students may succeed in “passing the exams” etc., they may not be able ot EXPLAIN these things or — a sure enough indicator fo understanding, — apply this complicated knowledge to solving complicated problems in real life. Just like famous mathematicians did and do…
    So you’re doing a great thing, thank you. I’m sure being able to EXPLAIN things is far more valuable than being an intuitive genius who’s unable to explain it to “uninitiated”.
    So thank you for your efforts, cause you bring me back to life. In former years (19th century) it used to be understood, that math is as important a part of human culture as literature. I agree it still is. Now if average human can read and understand books, why can’t he equally understand math? With your help it becomes possible. thanks again :))

  156. Hi Kostya, really glad it was clicking for you! Exactly, we need to look for the real-life applications and get a nice intuitive grounding [i.e., actually being able to explain the concepts]. And a large part is being able to convey our insights to others (what good does it do if the last loner math genius dies out?).

    I agree, math can be appreciated like literature, it’s well within our capabilities. Happy you’re enjoying the site :).

  157. Thank you for this explanation!

    I have been always wondering about the difference between the two and when to use them. I’ll remember your explanation for the rest of my life, I swear…

    -Joy

  158. Thank you so much for this wonderful article, it explains e and natural log pretty well. I may have to reread it a couple times for it to sink in but it’s a perfect explanation. One other thing that has been bothering me not just of e but of other functions like sin, cos, tan, pi, etc, is how their derived, from the view of someone plugging numbers into a calculator it’s just some magical black box that spits out a number. It’s much more then that, e for example can be defined as the function $$exp(x)=\frac{x^n}{n!} when x = 1 with a big enough n it converges to 2.71828.

  159. You are all belaboring the application of e to the subject matter of financial interest. This is only a tiny part of what e is. You think that you have a “deeper understanding” of this mathematical subject by going over many specific examples of one narrow application. But the only way to gain a truly “deep” understanding of any mathematical subject is to understand it and be able to work with it in its most general, compact and “terse” form (look up that last word if you don’t know what it means.. it is the hallmark of mathematics).

  160. I’ve looked up ln many, many times of the years and the meaning has always slipped right out from under me. Although I’m well-educated, this is just one of those holes in learning – something that was never taught properly to me in classes (through a doctorate level) or textbooks. This is hands-down the best and simplest explanation I have ever encountered. Not only do I understand seamlessly but I can now implement natural log and e^x as my work dictates.

  161. Hi sir,
    i have doubt in finding the variable ‘x’ in equation 1.5ln(x/2)+1/x=5.
    please clarify my doubt.
    Thank you.

  162. Again, a great article. While reading your articles, the emotions are quite mixed. On one side I am glad I finally understood the basics of logarithms. On the other hand, it makes me angry when I remember how I was taught these when in college (a really long time ago :)). It was just definitions followed by some dry theorems and of course, along with the so called rigorous proofs. It was all mechanical and memorization of formulae, nothing more. Absolutely no intuition. Your views are refreshing and my interest is renewed in Math again :-) and the need to look at math learning differently.

  163. Hey Kalid–

    Thanks to you, I’ve been getting comfortable with calculus, enough to move on to the Taylor series. But I have a question about a well-known pronouncement: Why is e^x considered to be the most important function in calculus?

    I know that e is a critical number, showing up almost everywhere, and I know that e^x is its own derivative. But what else does it do? How does it attain such a lofty status as the most important function in calculus?

  164. @Kumar: I know what you mean :). I’m simultaneously elated and frustrated when things click — why couldn’t it have been explained in simple terms the first time? All we can do is try to clear the path for other people.

    @Satheesh: Try http://betterexplained.com/articles/developing-your-intuition-for-math/, there’s an explanation of the 1/x property there.

    @Tim: Awesome, glad things are buzzing along. Great question on the role of e, try giving the above article (developing math intuition), it walks through several properties.

    Basically, e represents the idea of “continuous growth”. This is a very general principle which shows up almost anywhere you look. For example, a circle is “continuous rotation” and e^x can model a circle (with the right inputs). Most naturally growing things (bacteria colonies, radioactive decay, etc.) are changing continuously, instant-by-instant, so e^x is a nice way to describe how they change.

    From a calculus point of view, e^x is the *only* function which is its own derivative (sine/cosine also have this property, but under the covers, they are actually variations of e^x). Because of this, it’s very easy to analyze, and many tools of calculus, like the Fourier Transform, try to turn a given pattern into some variation on e^x. It’s like turning a Roman Numeral (XII) into decimal (12) so it’s much easier to multiply/add. If you can get your function in terms of e^x, it’s much easier to integrate/differentiate.

    In a nutshell, I might say this: Circles are essential to geometry and trigonometry. They’re well-studied, and if you can describe a pattern in terms of circles, you get all sorts of results “for free”. e^x is essential to calculus because it’s so well studied, and you get integration/derivatives “for free”.

  165. I am not a mathematician – a person not interested but life and career in finance led me into your world, after reading your articles I feel confident to explain to other people about e and logs. Thanks for your teaching talent.

  166. Thanks Moses, really glad you enjoyed it. One of my goals when writing is to help people experience concepts in a way they want to share with others themselves, and it’s very gratifying to hear when that’s the case.

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LaTeX: $$e=mc^2$$