# log wiki question

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• Oct 29th 2009, 06:27 PM
phish
log wiki question
The following formula comes from the wiki here on the constant "e"

http://upload.wikimedia.org/math/f/1...365ad2c211.png

where they substitute u=h/x in the last step for the 3rd equation. i've been grinding away, but can't figure out how they ended with that they went from the second equation to the third...i feel like i'm missing something obvious... any help would be greatly appreciated!http://en.wikipedia.org/wiki/E_%28ma...nt%29#Calculus
• Oct 30th 2009, 04:06 AM
Grandad
Hello phish

Welcome to Math Help Forum!
Quote:

Originally Posted by phish
The following formula comes from the wiki here on the constant "e"

http://upload.wikimedia.org/math/f/1...365ad2c211.png

where they substitute u=h/x in the last step for the 3rd equation. i've been grinding away, but can't figure out how they ended with that they went from the second equation to the third...i feel like i'm missing something obvious... any help would be greatly appreciated!http://en.wikipedia.org/wiki/E_%28ma...nt%29#Calculus

Using the laws of logarithms in the numerator:

$\lim_{h \to 0}\frac{\log_a(x+h) -\log_a(x)}{h}=\lim_{h \to 0}\frac{\log_a\left(\dfrac{x+h}{x}\right)}{h}$
$=\lim_{h \to 0}\frac{\log_a\left(1+\dfrac{h}{x}\right)}{h}$
Now when we substitute $u = \frac{h}{x}$, note that, in the context of this limit, it is $h$ that is the variable and $x$ a constant. So as $h \to 0, u \to 0$; and, of course, $h = ux$. So we get:
$=\lim_{u \to 0}\frac{\log_a(1+u)}{ux}$
and, taking out the constant $x$ from the limit gives us the desired result:
$=\frac{1}{x}\Big(\lim_{u \to 0}\frac{1}{u}\log_a(1+u)\Big)$
Grandad