logarithmic equation $\log_2(3^x-8)=2-x$ I've been unable to solve the following equation:
$$\log_2(3^x-8)=2-x$$
I can arrive at $$3^x-8=2^{2-x}$$ but I'm clueless afterwards. I know that the answer is $x=2$ but cannot arrive to that analytically. Thank you for any hint.

Let $f(x)=3^x-8$ and $g(x)=2^{2-x}$.
Since $f$ is an increasing function and $g$ is a decreasing function, 
our equation has one root maximum. 
But $2$ is a root, which gives an answer: $\{2\}$.
