Demystifying the Natural Logarithm (ln)

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.

[…] The time needed to grom from 1 to A is the time from 1 to 2, 2 to 3, 3 to 4… and so on, until you get to A. The first definition defines the natural log (ln) as shorthand for this “time to grow” computation. […]

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

[…] For example, the explanation of natural log is excellent. […]

Mike Y, you are a stupid, stupid DlCK.

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.

hey is there any laws dealing with natural logs over natural logs e.g ln|5 + 2x| / ln|5|?

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

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

assuming 100% growth rate, simplified statement?

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

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

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?

[…] Exponents & Logs: Intuitive exponents, natural log […]

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.

@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.

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!

@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 :).

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.

@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 :).

[…] site has excellent explanations of exponential and logarithmic functions too; the above link is to a page on calculus. And once again, as I read this stuff, I […]

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