The logarithm log has these two, seemingly very different, properties:
What's the connection?
You can prove the first from the second. And you don't even need to muck around with integrals and equations. The same proof works for any base, not just e; it's just that the diagrams are neatest for the natural logarithm.
ln(x) is the area underneath the curve of the graph f(x)=1/x from 1 to x. The following beautiful ASCIIvision should make everything "clear":
1 ++-----+-----+------+-----+------+-----+------+----++
** + + + + + + y = 1/x +
** |
*** |
**** |
0.8 +*** ++
***** |
****** |
******* |
******** |
0.6 +******** ++
*********** |
************* |
*************** |
****************: |
****************:::: |
0.4 +***************::::::: ++
****************::::::::::: |
****************::::::::::::::::: |
****************:::::::::::::::::::::::: |
****************::::::::::::::::::::::::xxxxxxxxx |
0.2 +***************:::::::::::::::::::::::: xxx|
****************:::::::::::::::::::::::: |
****************:::::::::::::::::::::::: |
****************:::::::::::::::::::::::: |
****************:::::::::::::::::::::::: |
0 +******+*****+******+*****+******+*****+******+*****+
1 1.5 2 2.5 3 3.5 4 4.5 5
The area filled with
* is the logarithm of a (=2 1/8). The combined areas filled with
* and
: are the logarithm of a⋅b (=2 1/8 * 1 15/17 = 4). What's the area filled with
:s?
Well, it's the area under the curve 1/x, from a to a⋅b. Your calculus textbook will probably have a formula or two for converting this to a more log-like formulation. But throw away your textbook, in favour of pretty pictures! Whom are you going to trust: Two tenured professors, or something you read on the Internet?
If we look at the curve a/x from a to ab, we have a curve that goes from a/a=1 to a/(ab)=1/b, over a distance of ab-a = a(b-1). So it's really just the curve for ln(b), stretched out a times wider (from a width of b-1 to a width of a(b-1)). Making a curve higher or wider by a factor of a increases the area under it by that factor. So the curve a/x from a to ab encloses an area of a⋅ln(b), and the curve 1/x from a to ab encloses an area of ln(b).
This proves that
ln(a⋅b) = ln(a) + ln(b).
More Useless Information
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The same argument also works in the other direction: If we want a function with the property of log, and we want it to be defined by an integral, very similar pictures to the above (hint: take b=1+δ/a) suffice to show that it must be a constant multiple of the integral of dt/t from 1 to x.
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Further up the mathematical pomposity scale, we've shown that log is an isomorphism of the multiplicative group ((0,∞),*) with the additive group (R,+). This is just a fancy-shmancy way of saying log(ab)=log(a)+log(b). But it also explains why slide rules work!
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John Napier didn't know about isomorphisms or integrals. He just wanted a trick to convert multiplication (which is hard) to addition (which is easy(ier)). Did he really have no idea that he was onto something so profound?
ASCIImation by Gnuplot ("set term dumb"), retouched by hand in XEmacs' picture-mode.