I'd like to prove that for $a,b>0$, $$(a+b)^x=a^x+b^x \implies x=1.$$
I tried to study the variation of the map $f:x\mapsto (a+b)^x-a^x-b^x$ but unfortunatly it's not monotone since $$f'(x)= (\ln (a+b)-\ln (a))a^x + (\ln (a+b)-\ln (b))b^x .$$ Taking $\ln$ in the equation doesn't seems to help either.
I feel like I'm missing something obvious. Any help will be greatly appreciate.
$(a+b)^x=a^x+b^x$ equivalent $1=(\frac a {a+b} )^x+(\frac b {a+b} )^x $
Because $0 \lt \frac a {a+b} \lt 1, 0 \lt \frac b {a+b} \lt 1$, the right side is strictly decreasing, therefore the solution $x=1$ is unique
You can do this even more simply: let $a=b=1$. Then on the left, $2^x$, and on the right, $1+1$.
EDIT: Though I may have misunderstood your question. I answered it as if you asked "If this holds for all $a,b$, then show that $x=1$". You may be asking instead "If this holds for some $a,b$, then show that $x=1$".
