Can someone explain why $(e,1)$ and $(t, \ln t)$ are the two points of intersection for this question? I was just going through Khan academy and this question completely threw me. I've rewatched the prior videos a few times to try to understand what I'm suppose to do, but I still don't understand.
The writing below the answer box is suppose to explain the question, but I'm stuck on why the points of intersection are $(e,1)$ and $(t, \ln t)$? I have no idea where that $1$ came from.

 A: $\ln$ is usually defined as a type of logarithm, the natural logarithm whose base is a constant called $e$.
$\ln e$ is defined to be the power you need to raise the special number $e$ to, to get $e$. Since $e^1=e$, $\ln e=1$.
A: It seems like the stumbling block here is that $e$ is not a variable.  It is a constant (like $\pi$), and it has a value of approximately $2.71828$.  It is, by definition, the base of the natural logarithm function $\ln$, and so
$$
\ln e \equiv \log_e e = 1
$$
for the same reason that $\log_{10} 10 = 1$, or $\log_2 2 = 1$, or $\log_\text{anything} \text{anything} = 1$: because if you raise anything to the first power, you just get that same anything.
Now, there may be a separate question here, which is why $e$ is that value and not some other value.  With $\pi$, we have a very simple geometric interpretation: It is the ratio of a circle's circumference to its diameter, no matter how big the circle is.
Unfortunately, $e$ does not have quite that simple a geometric interpretation.  Its uses are a bit more numerical.  Here's one that's not too involved.  Suppose you have money that earns interest at the rate of $1$ percent per year.  If you were to let that compound for $100$ years (so that $100$ times $1$ percent is just $1$), you would have approximately $e$ times the money you started off with.  (Of course, it would be worth a heck of a lot less due to inflation, but let's not concern ourselves with that here.)
It's only approximate because we've broken it up into only $100$ steps.  If we were to have it earn $1/12$ of a percent of interest every month, and let that run for $1200$ months ($100$ years, so that again $1200$ months times $1/12$ of a percent equals just $1$), it would be even closer to $e$ times your original investment.  And in the limit of continuous compounding (do banks ever advertise that anymore?), it would be exactly $e$ times your original amount.  (Well, up to rounding error.)
This is another way of saying that
$$
\left(1 + \frac{1}{n}\right)^n \doteq e \doteq 2.71828
$$
when $n$ is large, with the first equality becoming exact as $n$ increases without bound ("to infinity", in other words).  For instance, for $n = 100$, each year, the money becomes worth $1+1/100 = 1.01$ times the amount it was worth the previous year.  So after $100$ years, it's worth $1.01^{100}$ the amount it was originally.  The above formula says that that value is very close to $e$.  (The actual value is closer to $2.70481$.)
In fact, that is one way (there are others) to define the constant $e$:
$$
\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e
$$
Anyway, the definition of $e$ is not important to the present question, but considering that you're learning about derivatives (calculus derivatives, not financial ones), another definition that might interest you is the following: It is the unique constant that satisfies
$$
\frac{d}{dx} e^x = e^x
$$
If you take the derivative of any other exponential function, $b^x$, you don't get $b^x$, you get
$$
\frac{d}{dx} b^x = (\ln b) \, e^x
$$
Well, you get that for $e$ too, but since $\ln e = 1$, it works out.
