When $a^x=x,a>1$ Has Exactly One Solution

Observe that, for $a^x=x,a>1$ , if there is exactly one solution, then $a^x$ and $x$ are tangent to each other at the point of intersection.

Thus, $\frac{d}{dx}(a^x)=\frac{d}{dx}(x)$ , or $a^x\ln{a}=1$ . Solving this gives $x=\log_a{\frac{1}{\ln(a)}}=-\log_a{\ln(a)}$ .

Plug this in the original equation: $a^{-\log_a{\ln(a)}}=-\log_a{\ln(a)}$ , or $\frac{1}{\ln{a}}=-\frac{\ln{\ln(a)}}{\ln{a}}$ .

Solving this reveals that $a=e^{e^{-1}}$ . Thus, $x=-\log_a{\ln(a)}=e$ .

The final answer is $a+x=e^{e^{-1}}+e≈4.1629$ .