Geometric intuition for why the slope of the tangent at the y-intercept of $y=b^x$ is $\ln(b)$?

Question

I understand how to prove that algebraically, but it's really amazing how the slope is exactly $\ln(b)$. My question is how can I develop a geometric feeling for it, or is it even possible to do so?

Answer 1

If slope is $\frac {\Delta y}{\Delta x}=\frac {b^{x_1}-b^x}{x_1 - x}$ where $x_1$ is really really close to $x$.

If we let $x_h = x+h$ and $h$ is a really really small number then

slope is $\frac {b^{x+h} - b^{x}}{(x+h)-x} = \frac {b^xb^h -b^x}h= b^x \frac {b^h -1}h$

The value $\lim_{h\to 0^+} \frac {b^h-1}h$ is a constant value depending only on the value of $b$ and has nothing to do with $x$.

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Now how surprised would you be if I told you $\lim_\limits{h\to 0^+} \frac {b^h-1}h = \ln b$?

.... okay, maybe that is surprising.

But consider let define the function as $\operatorname{BEATSME}(b) = \lim_\limits{h\to 0^+} \frac {b^h-1}h$.

We can not that for small values of $b$ that $\operatorname{BEATSME}(b)$ is small and for large values of $b$ that $\operatorname{BEATSME}(b)$ is large.

So there must be some number where $\operatorname{BEATSME}(b) = 1$.

So lets define a number $E$ to be DEFINED to be the number where $\operatorname{BEATSME}(E) = 1$.

Would it surprise you if I told you that $E = e$ where $e$ is Euler's number?

Okay, most classes define $e$ as $\lim_\limits{n\to \infty} (1+\frac 1n)^n$. Okay... if we define $e$ that way what is $\operatorname{BEATSME}(e)$?

$e=\lim_\limits{n\to\infty} (1+\frac 1n)^n = \lim_\limits{h\to \infty}(1 + h)^{\frac 1h}$.

And $\operatorname{BEATSME}(e) = \lim_{h\to 0} \frac {e^g-1}h = \lim_\limits{h\to \infty}\frac {(\lim_\limits{h\to \infty}(1 + h)^{\frac 1h})^h-1}h=$

$\lim_\limits{h\to 0^+} \frac {((1+h)^{\frac 1h})^h-1}h=$

$\lim _\limits{h\to 0^+}{(1+h)^1 - 1}h = \lim_\limits{h\to 0^+} \frac {(1+h)-1}h=\lim_\limits{h\to 0^+} \frac hh = 1$.

So that's it . $\operatorname{BEATSME}(e) = 1$.

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Okay, but what is $\operatorname{BEATSME}(b)$ in general?

Okay $\operatorname{BEATSME}(b) = \lim_\limits{h\to 0} \frac{b^h -1}h= \ln b$.

Why?
Let $e^K = b$.

Then $\lim_\limits{n\to \infty} (1+\frac 1n)^{nK} = b$.

Let $(1+\frac 1n)^{nK} =B$ for a very large $n$.

Then $(1+\frac 1n)^K = B^{\frac 1n}$

By binomial expansion $(1+\frac 1n)^K \approx 1 + K\frac 1n$ for large $n$.

So $1+K\frac 1n = B^{\frac 1n}$

$K = n(B^{\frac 1n}-1)$.

Replace $n$ with $\frac 1h$ and we get $K = \frac {B^h-1}h$ and so the the

$\lim_\limits{h\to 0^+} \frac {b^h -1}h = \ln h$.

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Whew. So the slope of the tangent line of $b^x$ is $b^x\lim_\limits{h\to 0} \frac {b^h-1}h = b^x\ln b$.

And at $x=0$ that is $\ln b$ exactly.