Proving $\log(b^a) = a \log(b)$ using calculus Sorry, this is a really simple question, but I'm trying to teach myself calculus and can't figure it out. 
If we define $\log(b) = \frac{db^x}{dx}(0)$ how does one prove $\log(b^a) = a\log(b)$? I tried using the definition of derivative but got stuck at $\lim_{h \to 0}\frac{b^{ah} - 1}{h}$.
If $a\not=0$, let $k=ah$ and note that $h\to0$ iff $k\to0$, which gives

$$\lim_{h\to0}{(b^a)^h-1\over h}=a\lim_{h\to0}{b^{ah}-1\over ah}=a\lim_{k\to0}{b^k-1\over k}$$

If $a=0$, then $(b^a)^h-1=0$, so the limit is obviously $0$.
$$
\log(a^b) = \int_1^{a^b} \frac{dx}x.
$$
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,
$$
\log(a^b) = \int_1^{a^b} \frac{dx}x = \int_1^a b \frac{dw}w = b\log a.
$$