How do you prove that ln(a^x) = xln(a) and a^x = e^xln(a) without using exponent rules?

[logicgate]

Homework Statement

The proof must contain all real numbers x, not just for integers, it must be true for rational and irrational numbers.

Relevant Equations

ln(a^x) = xln(a), a^x = e^xln(a)

In the book "Calculus by Michael Spivak" it says that a^x = e^xln(a) is a definition. And I am not convinced to accept this as true without a proof.

 

[fresh_42]

You want to define $a^x$ for any real number $x$. I suppose we can assume $a\geq 0$ for otherwise, we would get into trouble already with rational values of $x.$ I know what $a^z$ for $z\in \mathbb{Z}$ is since we can define it recursively. That allows us to define $a^{1/n}=\sqrt[n]{a}$ since an $n$ long product of them yields $1.$ We get therefore the definition for all $x\in \mathbb{Q}.$

Now, it gets complicated. The real numbers are limits of certain sequences of rational numbers. Say we have $x=\displaystyle{\lim_{n \to \infty}}x_n$ and we require that $x \mapsto a^x$ is continuous. We need this condition so that we can define $a^x= \displaystyle{\lim_{n \to \infty}}a^{x_n}.$ The requirement makes sense if we look at the drawings of $x_n \mapsto a^{x_n}.$ However, it is an artificial requirement in the sense that we can only get from the rational to the reals if we accept a topological homomorphism, a continuous function.

Thus, we have defined $a^x$ by algebraic and topological means, that mimic the process from natural numbers over integers and rational numbers to real numbers.

So far so good. Now, what are $x\mapsto e^x$ and $a\mapsto \ln(a)$? You have to define them before we prove the statement. For example, their definition by functional equations is possibly less helpful than their definition by power series or limits. So what are they to you? In order to prove something, we must agree on what we are allowed to use.

 

[logicgate]

use power series

 

[fresh_42]

Let us use that $e^{\ln(x)}=x$ or prove it with your power series. With that, we only have to show that $x\cdot \ln a = \ln a^x$ and the second equation follows. We can also assume that $x>0$ since the negative case ...
\begin{align*}
\ln a^{-x}&=\ln \dfrac{1}{a^x}=\ln \left(\dfrac{1}{a}\right)^x =x\ln \left(\dfrac{1}{a}\right)\\
&=x\cdot \left(\ln 1 -\ln a\right)=x\cdot(0-\ln a)=-x\ln a
\end{align*}
... follows from the positive case and $x=0$ is trivial. I would suggest that you use
$$\ln a^x=\lim _{h\to 0}\int _{1}^{a^x}{\frac {1}{t^{1-h}}}\,\mathrm {d} t$$
as definition and substitute $s=t^{1/x}.$

 

[logicgate]

How did you arrive to the conclusion that ln(1/a)^x = xln(1/a) ? This is exactly what I want to prove.

 

[fresh_42]

I did not arrive there. I proposed possible paths. You could prove it for rationals and use continuity and the fact that $\mathbb{Q}\subseteq \mathbb{R}$ is dense, or you can use the integral definition and the substitution I suggested.

However, you should do some of the steps on your own.

 

[jack action]

By definition:
$$b^n = \underbrace{b \times b\ ... \times b}_{n \text{ times}}$$
Therefore we get our first identity:
$$b^{n+m} = \underbrace{b \times b\ ... \times b}_{n \text{ times}}\times\underbrace{b \times b\ ... \times b}_{m \text{ times}}$$
$$b^{n+m} = b^n b^m$$
Similarly, we can get our second identity:
$$\left({b^n}\right)^m = \underbrace{\underbrace{b \times b\ ... \times b}_{n \text{ times}}\times\underbrace{b \times b\ ... \times b}_{n \text{ times}}\ ... \times \underbrace{b \times b\ ... \times b}_{n \text{ times}}}_{m \text{ times}}$$
$$\left({b^n}\right)^m = b^{nm}$$
About how this applies to rational and irrational numbers you can read the Wikipedia page about exponentiation starting here.

If you accept that and we define that $y = \log_b x$ if $b^y = x$, then by our first identity:
$$b^{y_1}b^{y_2} = b^{y_1 + y_2}$$
$$x_1x_2 = b^{y_1 + y_2}$$
$$\log_b(x_1x_2) = y_1+y_2$$
$$\log_b(x_1x_2) = \log_b x_1+\log_b x_2$$
From there, if $x_1 = x_2 = a$, then:
$$\log_b(a^2) = \log_b a+\log_b a = 2 \log_b a$$
And for $\log_b(a^3)$, we get:
$$\log_b(a^2\times a) = 2\log_b a+\log_b a = 3 \log_b a$$
Or for the general case:
$$\log_b(a^x) = x \log_b a$$
As for the other case, we used the second identity:
$$a^x = \left(a\right)^x$$
$$a^x = \left(e^{ln(a)}\right)^x$$
$$a^x = e^{xln(a)}$$