A Special Property of e? I was debating whether I should post this question over at MathEducatorsStackExchange or here, sorry if this was the wrong forum for this question. 
I am teaching Calculus I this semester and I found the following problem, which I gave to a group for one of their group challenge problems:
"Assume $b$ is a positive number and $b≠1$. Find the $y$-coordinate of the point on the curve $y=b^x$ at which the tangent line passes through the origin."
The group was able to solve the problem (the $y$-coordinate turns out to be $b^{\frac{1}{\ln(b)}}$ which is the same as $e$). They also created a Desmos graph to demonstrate this property.
However, they asked the question: "why is this true?" They know that the steps of the problem show it to be true, but they wanted to know if there was some deeper property of $e$ that made this property more obvious, or if there was some intuition as to why this property exists. Or, is this just a neat Calculus problem with a tidy answer?
EDIT: For some clarification, the students weren't asking "why is $b^{\frac{1}{ln(b)}}=e$", they were able to figure that out after I suggested they plug in $b=2, 3, 4,...$ etc. and see if they notice a pattern. They were more asking about "why is the tangent line at $b^x$ that goes through the origin always tangent at the point $(x, e)$. My initial response was to look at there work, they had supplied the answer to that question just by doing the problem (like some people are suggesting in the comments). They seemed unsatisfied though, and thought there was some intuition that might make this property more obvious.
 A: I think the first point is that we can remove a parameter.  Consider the transformation: $x \to c x$, $b \to b^{1/c}$ where $c > 0$.  This just stretches the picture in the horizontal direction by changing $b$.
For an appropriate choice of $c$ we can get $b=e$.
Then the rest is the fact that $y = e^x$ has derivative $e$ at $x=1$.
A: The special property is
$$(e^x)'=e^x.$$
For any base,
$$(b^x)'=cb^x$$ and $e$ is the only one for which $c=1$. This is why it is called the natural base and why it is so widespread in calculus. (For a similar reason, angles are counted in radians.)
A: I think the lynchpin here, which sets $e$ apart from the rest of the real numbers, is that $e$ is the base of the natural logarithm. In other words, the constant $\ln b$ which appears when you differentiate $b^x$ is what connects this to $e$ more than any other number.
Even without knowing anything about the connection between logarithms and differentiating exponential functions, and just using that the derivative of $b^x$ is $cb^x$ for some number $c$ (which amounts to showing that $c = \lim_{h\to 0}\frac{b^h - 1}{h}$ is a limit which exists), you will find that the point of tangency for the relevant tangent must be $(\frac1c, b^{1/c})$.
A: I. Why $b^{\frac{1}{\ln(b)}}=e$- note 2 logarithm properties:


*

*$\frac{1}{log_p(q)}=log_q(p)$

*$B^{log_B(C)}=C$
II. The history and (quick) winning of $e$:
Find the derivative of the function $f(x)=a^x$ by definition:
$$\displaystyle \lim_{h\rightarrow0}\frac{a^{x+h}-a^x}{h}=\displaystyle \lim_{h\rightarrow0}\frac{a^x a^h-a^x}{h}=\displaystyle \lim_{h\rightarrow0}\frac{a^x (a^h-1)}{h}=a^x\cdot\displaystyle \lim_{h\rightarrow0}\frac{a^h-1}{h}
$$
As we see- the derivative of $a^x$ is $a^x$ multiple by this something.
Euler so "For which $a$ is the rest equal $1$?"
Hence (in quick):
$$\displaystyle \lim_{h\rightarrow0}\frac{e^h-1}{h}=1
$$
here a fact: $something_{tiny}=\frac{1}{something_{big}}$
$$\displaystyle \lim_{n\rightarrow\infty}\frac{e^{\frac1n}-1}{\frac1n}=1
$$
Now we ignore for a minute the "$\displaystyle \lim_{n\rightarrow\infty}$":
$$\frac{e^{\frac1n}-1}{\frac1n}=1
$$
$$e^{\frac1n}-1=\frac1n
$$
$$e^{\frac1n}=1+\frac1n
$$
$$e=(1+\frac1n)^n
$$
$$\bbox[5px,border:2px solid blue]{e=\displaystyle \lim_{n\rightarrow\infty}(1+\frac1n)^n}
$$
A: “Why is the tangent line at $b^x$ that goes through the origin always tangent at the point $(x, e)$?”
Consider a slightly more general case where $b^{x_0}=e^a$, that is, $x_0=a/\ln(b)$.  One easily calculates the $y$-intercept of the tangent in $x_0$ as $b^{x_0}\bigl(1-\ln(b)x_0\bigr)$.  Now if $b^{x_0}=e^a$ it becomes $e^a(1-a)$, independently of $b$.
Therefore there’s an easy way to construct the tangent.  Draw $e^x$ and $b^x$ in the same coordinate system.  Let $E(a,e^a)$ and let $B$ be the point on $b^x$ with the same second coordinate.  Then the tangents is these points will meet in $(0,1-a)$.
I didn’t know this little cutie before, so thanks a lot.
