Proving $\log(b^a) = a \log(b)$ using calculus

Question

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 $\lim_{h \to 0}\frac{b^{ah} - 1}{h}$.

Answer 1

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 $0$.

Answer 2

Hint: $\displaystyle \log(b^a) = \frac{d(b^{ax})}{dx}(0)$ and using the chain rule $$\frac{d}{dx}b^{ax} = \frac{d}{dx}(b^{x})^a = a(b^x)^{a-1}\frac{d}{dx}b^{x}$$

Answer 3

Using $(b^a)^x = b^{ax}$ and the chain rule, we get \begin{align} \log(b^a) &= \left.\frac{\mathrm d}{\mathrm dx} b^{ax}\right|_{x=0} = \left.\left(\frac{\mathrm d}{\mathrm dy} b^y\right)\right|_{y=0} \cdot\left.\left( \frac{\mathrm d}{\mathrm dx} ax\right)\right|_{x=0} = \log(b)\cdot a. \end{align}

Answer 4

$$ \log(a^b) = \int_1^{a^b} \frac{dx}x. $$ 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, $$ \log(a^b) = \int_1^{a^b} \frac{dx}x = \int_1^a b \frac{dw}w = b\log a. $$