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 $a(b)$.