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

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 $$\{ c \in \mathbb{R} \mid 2 such that $$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?

• It really depends on how you define the natural logarithm function. Is it defined as an antiderivative of $g(x) = 1/x$? As the inverse of the natural exponential, $f(x) = e^x$ (and if so, how is this defined?)? If you want to prove two things are identical you have to first have an independent definition of what each one is, so that you can prove they are equivalent. Commented Dec 30, 2020 at 19:47

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