Prove $L(b) = \ln(b)$

Question

While thinking about the Taylor expansion of $b^x$, to take the derivative of $b^x$ we need to find: $$L(b) = \lim_{h \to 0} \frac{b^h-1}{h}$$

It can be shown numerically that $L(2) \approx 0.7 $ and $L(3) \approx 1.1 $ thus by the Intermediate Value Theorem there must be a $ \{ c \in \mathbb{R} \mid 2<c< 3\}$ such that $L(c) = 1 $. Of course, we call that value Euler's number or e.

I have heard it asserted that $L(b) = \ln(b)$ but am unsure how to derive that result?

Answer 1

The question finds an immediate answer by noting that $b^h=e^{\ln b^h}=e^{h\ln b}$. Therefore $$L(b) = \lim_{h\to 0}\frac{b^h-1}{h}=\lim_{h\to 0}\frac{e^{h\ln b}-1}{h} = \ln b\lim_{t\to 0}\frac{e^t-1}{t}=\ln b.$$ I have multiplied and divided by "$\ln b$" the fraction and substituted $h\ln b = t$ to get the result.

Answer 2

Hint:

$$L(b)=\frac{d}{dx}\left(b^x\right)\mid_{x=0}$$

Answer 3

One approach is to define $e$ as the unique number $b$ for which $$ \lim_{h \to 0}\frac{b^h-1}{h}=1 \, . $$ It then directly follows that $e^x$ is its own derivative. Then, since $b^h=e^{h\log(b)}$, it can be shown that the above limit is equal to $\log(b)$ in general. (See Gauge_name's answer.)