How to solve $a^x=a+x$ Given the equation $a^x = a + x$, how do you express $x$ in terms of $a$? 

This equation has a solution expressed in terms of Lambert function and the solution is $$x=-a-\frac{W\left(-a^{-a} \log (a)\right)}{\log (a)}$$ In fact, any equation which can be written as $$A+Bx+C\log(D+Ex)=0$$ has explicit solution(s) in terms of Lambert function.
In the real domain, $W(t)$ only exists if $t \geq -\frac{1}{e}$. For your problem, the only restriction seems to be $a>0$.
In order to approach it, you can write $$a^x=\frac{a^{x+a}}{a^a}$$ and so the equation becomes $$a^{x+a}=e^{(x+a)\log(a)}=(x+a)\log(a)\frac{a^a}{\log(a)}$$ and define $$y=(x+a)\log(a)$$ which make the equation $$e^y=y\frac{a^a}{\log(a)}$$ or still better $$ze^{z}=-\frac{log(a)}{a^a}$$ where $z=-y$.
