TITLE: Show that there is a simultaneously tangent line to both the curves $y = e^x$ and $y = \ln x$
QUESTION [6 upvotes]: I found the derivatives of both of the curves but I'm having trouble on how to move on from here:
$y = e^x\implies y' = e^x$
$y = \ln x\implies y' = \frac{1}{x}$

REPLY [4 votes]: Let the tangent have the equation $y=ax+b$.
The condition of tangency to the exponential at $x_e$ is written
$$ax_e+b=e^{x_e},a=e^{x_e}$$
and that to the logarithm at $x_l$ is
$$ax_l+b=\log x_l,a=\frac1{x_l}.$$
We eliminate $x_e,x_l$,
$$a\log a+b=a$$
$$1+b=-\log a$$
then $b$, and we get
$$a\log a-a-\log a-1=0.$$
As the LHS is continuous, negative at $a=e$ and positive at $a=e^2$, there is certainly a solution, hence a common tangent.

As mentioned by @user3733558, there is another solution. By symmetry, it must correspond to the substitution $a\leftrightarrow\dfrac1a$. Indeed we verify
$$-\frac{\log a}a-\frac1a+\log a-1=\frac{a\log a-a-\log a-1}a=0.$$