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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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Those are not derivatives; they are just new names for the expressions. You could use $a=\exp(x)$, $b=\exp(y)$, so that $x=\log(a)$, $y=\log(b)$. Then $x+y = \log(a)+\log(b) = \log(ab)$ (presumably you already established the identities for logarithms); then taking exponentials you get $\exp(x+y) = \exp(\log(ab)) = ab = \exp(x)\exp(y)$. –  Arturo Magidin Nov 28 '11 at 4:53
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I think you want to say in the last sentence that $\exp(x+y) = x'y' = \exp(x)\exp(y)$? –  user38268 Nov 28 '11 at 5:23
    
@ArturoMagidin: I don't get the part $ \exp(\log(ab)) = ab $ , could you please explain it? –  Gigili Nov 28 '11 at 7:17
    
That should follow from exp being defined as $\log^{-1}$. –  Mike Nov 28 '11 at 7:56
    
@Mike: Oh yes, sorry for being silly. –  Gigili Nov 28 '11 at 8:43
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1 Answer 1

up vote 4 down vote accepted

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.

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