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

First note that if $a^x=x^a,a>1$ has exactly one solution then $a^x$ and $x^a$ is tangent to each other at the point of intersection. Further note that, the one solution must be $x=a$ .

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

Substitute $x=a$ to get $a^a\ln{a}=aa^{a-1}=a^a$ . Thus, $a^a(\ln{a}-1)=0$ . Since $a>1$ , $a^a≠0$ . The only solution is when $\ln{a}=1$ , or $a=x=e$ .

The final answer is $e-\ln{e^e}-\ln{(e\cdot e)}=-2$ .