I have found the following surprisingly nice result.
The graph of $y=a^x$ intersects its inverse $y=\log_a(x)$ when $a=e^{\frac{1}{e}}$ at $x=e$.
Question: Is there a faster way of proving this result than below? The method below feels too long.
My method: Let the intersection point be at $x=X$.
The two curves must intersect on the line $y=x$. $$a^X=X$$
The line $y=x$ is a tangent.
$$a^X \ln (a) = 1$$
Combine results.
$$X =\frac{1}{\ln (a)} \,\,\,\,\,\,\,(*)$$
But also
$$\log_a(X)=X=\frac{\ln(X)}{\ln(Aa)}$$
Substitute in $(*)$
$$\frac{1}{\ln(a)}=\frac{\ln(X)}{\ln(a)}$$
Hence $x=e$. $$a^e=e$$ Therefore $$a=e^{\frac{1}{e}} \,\,\,\,\,\, \square$$
The two graphs are inverses, thus their intersection points all happen when either graph passes through $y=x$.
Hence you are looking to solve: $$a^x=x$$ or$$\log_a(x)=x$$This is at $a=x^\frac1x$ or $a=\sqrt[x]{x}$ (same thing but second one is more obvious).