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Source: Spivak's Calculus Chapter 18: The Logarithm and Exponential Functions. Theorem 3:

Theorem: For all numbers $x$, $\exp(x+y)=\exp(x)\exp(y)$, where $\exp$ is defined as $\log^{-1}$

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Proof: Let x' = \exp(x) and y' = \exp(y), so that

x = \log x',

y = \log y'.

Then

x + y = \log x' + \log y' = \log(x'y').

This means that

\exp(x + y) = x'y' = \exp(x) \exp(y).

I don't get the beginning part where he lets x' = \exp(x) and y'=\exp(y)... Could he have used f'= \exp(x) and g'=\exp(y) for less confusion, or am I misunderstanding something completely?

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    @Mike: Oh yes, sorry for being silly.2011-11-28

1 Answers 1

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Those are not derivatives, they are just new names for the expressions.

To make it clearer, let's use completely new names. Here, $x$ and $y$ are fixed real numbers; let $a=\exp(x)$, and $b=\exp(y)$. Then, by the definition of exponential and logarithms, we have $x=\log(a)$ and $y=\log(b)$. Thus, $x+y=\log(a)+\log(b) = \log(ab)$ (presumably this property of logarithms has already been established). So $\exp(x+y) = \exp(\log(ab)) = ab = \exp(x)\exp(y),$ with $\exp(\log(ab))=ab$ because $\exp$ is the inverse function of the logarithm.