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I am trying to understand the definition of a logarithm, because when I was trying to find the derivative of $2^x$ I got $2^x \lim_{h \to 0} \frac{2^h-1}{h}$ which I have found by searching to be $\ln(2)$. I did get a bit confused because I would need to use l'hopital rule, which would bring be back to what I was trying to find.

But my question that I think I need to understand before getting to my second question.

Euler defines logarithm as $\ln(x)=\lim_{n \to \infty}n(x^{\tfrac{1}{n}}-1)$

Which then must be equal to $\ln(x)=\lim_{h \to 0} \frac{x^h-1}{h} $

Could you help me understand how these are both the same?

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    I posted [this related answer](http://math.stackexchange.com/questions/190773/proof-of-fracddxex-ex/190780#190780) here recently.2012-09-09

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These are both the same since if you define $h = 1/n$, you get your second equation from the first.

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If you are trying to find the derivative of $2^x$ then you also can do it using the property of $e$

$ \lim_{h \to 0} \frac{2^h-1}{h} = \lim_{h \to 0} \frac{e^{\ln(2) h}-1}{h \ln (2)}\times \ln(2) = \ln (2) \times 1 = \ln(2)$

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    @Arthur thanks alot2012-09-09