I have been trying to find the shortest distance between $f(x)=e^x$ and $g(x)=\ln(x)$.
All methods I have seen include taking the $y=x$ line as the mirror and find a point on both curves which have same slope as the mirror ($1$ in this case).
Now this method is working only because it is a very specific case as the functions are inverse of each other. The general method would be to find a common normal between the two curves and calculate the distance from where the normal intersects them.
I am trying to apply this method for the given question but am getting unexpected results.
First I found derivatives for both curves, took their reciprocals, and multiplied them by a minus sign to get the slope of the normal:
$m_{f(x)normal}=-\frac{1}{e^x}$
$m_{g(x)normal}=-x$
Now, both these slopes must be equal in case of a common normal.
But equating them yields:
$xe^x=1$
This is far from the actual answer of $x=1$ and $x=0$ where the normal intersect the curves.
Where am I going wrong?
The slopes you have computed for the normal for each curve are correct, but it is not always necessary that they are equal for the same value of $x$.
It must be the case that for some $x_1$, the slope of the normal at a point $(x_1, e^{x_1})$ on the curve $y=e^x$ is $-\frac{1}{e^{x_1}}$, which is equal to $(-x_2)$ which is the slope of the normal at another point $(x_2,\ln(x_2))$ on the curve $y= \ln x $.
Proceeding from this premise, we write the equations for the normal for $y=e^x$ and $y=\ln x$:
$$y-e^{x_1} = -\frac{1}{e^{x_1}}\cdot (x-x_1)$$
$$ y - \ln(x_2) = -x_2 \cdot (x-x_2)$$
Now, since these two equations represent the same normal, we equate the coefficients of $y$, $x$ and the constants. We obtain:
$$ 1 = 1$$
$$-\frac{1}{e^{x_1}} = -x_2 \tag1$$
and
$$ \frac{x_1}{e^{x_1}} + e^{x_1} = x_2^2 + \ln(x_2) \tag2$$
Using equation $(1)$, we rewrite rquation $(2)$ as:
$$ x_1 x_2 + \frac{1}{x_2} = x_2^2 + \ln(x_2)$$
From this, you can see that the values $x_1=0$ and $x_2=1$ satisfy the equations.