# Prove that, if $r \in \mathbb{Q}$, then $\mathrm{exp}_{a}(r) = a^r$

I am using this material to study calculus.

The logarithm (Sect. 2.2.4) has just been introduced as:

Given a number $$a > 0$$, $$a \neq 1$$, and a number $$x > 0$$, the logarithm of $$x$$ to the base $$a$$ is defined as the only number $$y \in \mathbb{R}$$ that verifies $$a^y = x$$.

The only property explicitly given for the logarithm is

$$\log_a(xy)= \log_a(x) + \log_a(y) (x>0, y>0).$$

and the formula for the change of base.

In the following section (2.2.5) the exponential function is introduced as the inverse of the logarithm function, so by definition

given $$x \in \mathbb{R}$$, $$\exp_{a}(x)$$ is the only positive number such that $$\log_{a}(\exp_{a}(x)) = x$$.

Then it says that

it is easy to prove that, if $$r \in \mathbb{Q}$$, then $$\exp_{a}(r) = a^r$$. Thus the notation $$\exp_{a}(x) = a^x$$ is used.

I cannot prove this. At first I thought this equality ($$\exp_{a}(r) = a^r$$) was implicit to the definition of the exponential (and the logarithm), but then I would not understand what is the relevance of mentioning rational numbers to prove it ($$r \in \mathbb{Q}$$).

I have found many other examples in math stackexchange, but they consider the base $$e$$ and often use the definition of $$e$$ as a limit or its power series representation to prove the equality $$\exp(x) = e^x$$. As this is just the second chapter of the book I feel I should be able to prove $$\exp_{a}(r) = a^r$$ for $$r \in \mathbb{Q}$$, only using the definition of logarithm and the property of the logarithm of a product.

A first, broken, attempt at proving it follows:

If $$r \in \mathbb{Q}$$, then it can be written as $$r = m/n$$ with $$m \in \mathbb{Z}$$ and $$n \in \mathbb{N}$$. By definition of exponential:

$$\log_{a}(\exp_{a}(r))= r = \frac{m}{n},$$

if we could also prove that

$$\log_{a}(a^r) = r = \frac{m}{n},$$

then we would have proved that $$\exp_{a}(r) = a^r$$, being the logarithm injective (this is a property already presented in this chapter of the book).

So

$$\log_{a}(a^r) = \log_{a}(a^{m/n}) = log_{a}(\underbrace{a^{1/n}a^{1/n}...a^{1/n}}_{\text{m times}}) = m \log_{a}(a^{1/n})$$

and here I don't know how to continue, since the property of the logarithm of a product does not tell me how to deal with rational powers. And I don't think I can directly simplify $$\log_{a}(a^{1/n}) = 1/n$$, otherwise I could have done that straight away and write $$\log_{a}(a^{r}) = r$$.

I think the approach is right because rational numbers facilitate the factorisation of the power and the subsequent application of the property of the log of a product.

• The definitions of "this material" are nonsense, or at best circular. What is "the only number $y \in \mathbb{R}$ that verifies $a^y = x$" if $a^y$ has been defined before only when $y\in\Bbb Q$? Better use a reliable textbook. That said, you can set $b=a^{1/n}$ and use that $\log_a(b^n)=n\log_a(b).$ Commented Jan 24, 2023 at 15:55
• I would argue that the author is defining (at least initially) $a^r$ as separate from $\exp_a(r)$. The question is how to prove they are in fact the same. Commented Jan 24, 2023 at 16:35
• @QC_QAOA did you have a look at "this material"? the author defines $a^r$ only for $r$ rational. Commented Jan 24, 2023 at 16:40
• Your right, thats pretty handwavy Commented Jan 24, 2023 at 16:44

As pointed out in the comments, your book seems to be lacking some details; specifically how to define $$a^r$$ for non-rational $$r$$. However, we can still make some headway on your question as it specifically asks about $$r\in\mathbb{Q}$$. To start, note that by induction
$$\log_a(x^n)=\log_a(x)+\log_a(x^{n-1})=\log_a(x)+(n-1)\log_a(x)=n\log_a(x)$$
for any $$n\in\mathbb{N}$$. Then
$$\log_a(a^r)=\log_a(a^{m/n})=\frac{n}{n}\log_a(a^{m/n})=\frac{1}{n}\log_a((a^{m/n})^n)=\frac{1}{n}\log_a(a^m)=\frac{m}{n}\log_a(a)=\frac{m}{n}=r$$
Thus, for $$r\in\mathbb{Q}$$ we have $$\exp_a(r)=a^r$$. Of course, what happens for $$r\in\mathbb{R}/\mathbb{Q}$$ is anyones guess at this point.