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In Chapter 18 of Spivak's Calculus, the author gives some exposition on the $\ln$ and $\exp$ functions; there is an accepted answer (Can you raise a number to an irrational exponent?) elsewhere that nicely summarizes Spivak's general approach. After Spivak remarks that one can show $\exp(x)=[\exp(1)]^x$ for any $x \in \mathbb Q$...and after defining $e^x=\exp(x)$ for any $x \in \mathbb R$, he writes the following:

The terminology "exponential function" should now be clear. We have succeeded in defining $e^x$ for an arbitrary (even irrational) exponent $x$. We have not yet defined $a^x$, if $a \neq e$, but there is a reasonable principle to guide us in the attempt. If $x$ is rational, then $a^x=(e^{\log a})^x\color{red}{=}e^{x \log a}$.

This final bolded sentence, in particular the equality colored $\color{red}{\text{red}}$, has me confused. It is not clear to me why this is true. I recently proved that if $x,y \in \mathbb Q$, then $(\alpha^x)^y=\alpha^{xy}$, but I have no proof that if $x \lor y \notin \mathbb Q$, then the statement remains true. Clearly, because $\log$ is a continuous function with an image of $\mathbb R$, it must be the case that there exists $a$ such that $\log a \notin \mathbb Q$. This means that I am not eligible to apply my $(\alpha^x)^y= \alpha^{xy}$ rule.

If $x$ is a non-zero integer, then I can prove that $(e^{\log a})^x=e^{x\log a}$ using the property that for any $r_1,r_2 \in \mathbb r: \exp(r_1+r_2)=\exp(r_1)\exp(r_2) \quad (\star_1)$. However, if $x$ is a rational number whose denominator is not $1$, then it is unclear to me how I resolve the root symbol.

For example, let $x =\frac{p}{q}$. Then, by definition, we have that $(e^{\log a})^{\frac{p}{q}}=\sqrt[q]{\left(e^{\log a}\right)^p}$. From the aforementioned property $(\star_1)$, this then equals $\sqrt[q]{e^{p \log a}}$, which, by definition, is just $\left(\exp(p \log a )\right)^{\frac{1}{q}}$. But how do I get the $q$ into the exponent...or, equivalently, how do I get $q$ in to the denominator of the argument for $\exp$?

S.C.
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1 Answers1

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I think we can use an altogether different approach...one that relies on the property that $\log$ and $\exp$ are inverse functions, which, importantly, means that they are injective functions!

We will require the following property of $\log$:

If $x,y \gt 0$, then $\log(xy)=\log(x)+\log(y) \quad (\star_1)$

With this in mind, if I can prove that $\log\left(\left(e^{\log a}\right)^x\right)=\log \left(e^{x\log a}\right)$ for $x= \frac{p}{q}$, then we must have that $\left(e^{\log a}\right)^x=e^{x \log a}$.

The right side is trivial. We have that $\log (e^{x \log a})=\log \circ \exp (x \log a)=x \log a = \frac{p}{q}\log a$

Now, for the left side, for easier reading, let us set $e^{\log a}=\alpha$.

Then we have that $\log(\alpha^x)=\log(\alpha^{\frac{p}{q}})=\log\left(\sqrt[q]{\alpha^p}\right)=\log\left(\underbrace{\sqrt[q]{\alpha}\times \cdots \times \sqrt[q]{\alpha}}_{p \text{ times}}\right)$

By repeatedly applying $(\star_1)$, we have that:

$$\log\left(\underbrace{\sqrt[q]{\alpha}\times \cdots \times \sqrt[q]{\alpha}}_{p \text{ times}}\right)=p\log\left(\alpha^{\frac{1}{q}}\right)$$

How should we deal with $\alpha^{\frac{1}{q}}$?

Well, note that $\log\left(\underbrace{x^{\frac{1}{q}}\times \cdots \times x^{\frac{1}{q}}}_{q \text{ times}}\right)=\log(x)$. But using $(\star_1)$, we also know that $\log\left(\underbrace{x^{\frac{1}{q}}\times \cdots \times x^{\frac{1}{q}}}_{q \text{ times}}\right)=q\log\left(x^{\frac{1}{q}}\right)$. This means that $\log(x)=q\log\left(x^{\frac{1}{q}}\right)$...or, equivalently, that $\frac{\log(x)}{q}=\log(x^{\frac{1}{q}})$.

Returning to our original proof, we thus have that $p \log\left(\alpha^{\frac{1}{q}}\right)=\frac{p}{q}\log(\alpha)=\frac{p}{q}\log(e^{\log a})=\frac{p}{q}\log \circ \exp (\log a)=\frac{p}{q}\log a$

S.C.
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