What is the solution of $a^b=a+b$ in terms of $a$? 
Let $a, b$ be real numbers. Solve
  $$a^b=a+b$$
  for $a$.

If there isn't a solution with $a, b$ real, maybe $a, b$ should be complex.
But no matter how hard I try, this is proving to be very difficult to
do. Would anyway be kind enough to show me the solution to this? Thank
you!

I will use Lambert-W function defined as the following:
$$W(x)e^{W(x)}=x$$
Your equation:
$$a^b=a+b$$
Multiply it by $a^a$ to make the exponent more "friendly" and use some tricks
$$a^{a+b}=(a+b)a^a$$
$$\frac{1}{a+b}a^{a+b}=a^a$$
$$(a+b)a^{-(a+b)}=a^{-a}$$
$$-(a+b)a^{-(a+b)}=-a^{-a}$$
$$-\ln(a)(a+b)e^{-\ln(a)(a+b)}=-\ln(a)a^{-a}$$
$$-\ln(a)(a+b)= W(-\ln(a)a^{-a})$$
$$a+b= -\frac{W(-\ln(a)a^{-a})}{\ln(a)}$$
$$b= -\frac{W(-\ln(a)a^{-a})}{\ln(a)}-a$$
When solving for $b$ is easy, solving for $a$ is not and I'm not sure if even possible using known functions.
EDIT: Here is the plot of $a(b)$.
