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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.

Thanks in advance.

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    $\begingroup$ 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 $\log_a(b^n)=n\log_a(b).$ $\endgroup$ Commented Jan 24, 2023 at 15:55
  • $\begingroup$ 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. $\endgroup$
    – QC_QAOA
    Commented Jan 24, 2023 at 16:35
  • $\begingroup$ @QC_QAOA did you have a look at "this material"? the author defines $a^r$ only for $r$ rational. $\endgroup$ Commented Jan 24, 2023 at 16:40
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    $\begingroup$ Your right, thats pretty handwavy $\endgroup$
    – QC_QAOA
    Commented Jan 24, 2023 at 16:44

1 Answer 1

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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.

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