Proof that $\log(a^b) = b\log a$ when $\log$ is defined by an integral

Question

When $\log a$ is defined as $\displaystyle\int_1^a\frac{dx}x$, then how does one prove that $\log(a^b)=b\log a$?

I will post an answer here that is identical to an answer I posted to another question. The original poster down-voted it saying that's not the definition of logarithm he was using.

Answer 1

$$ \log(a^b) = \int_1^{a^b} \frac{dx}x. $$ Suppose $w^{\,b}=x$. Then $bw^{b-1}\,dw= dx$, so $b\dfrac{dw}{w} = \dfrac{dx}x$. As $x$ goes from $1$ to $a^b$, then $w$ goes from $1$ to $a$. Hence, $$ \log(a^b) = \int_1^{a^b} \frac{dx}x = \int_1^a b \frac{dw}w = b\log a. $$

Answer 2

Alternative proof

Let

$$F(a)=\int_1^{a^b}\frac{dx}{x}$$ then by the fundemental theorem of analysis $$F'(a)=\frac{1}{a^b}\times ba^{b-1}=\frac ba$$ so we integrate we find

$$F(a)=b\log a+C$$ and using that $F(1)=0$ we conclude that $$F(a)=\log(a^b)=b\log a$$