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.

$\log_2(3^x-8)=2-x$$\implies 3^x-8=2^{2-x}$
For all value of $x$ in natural number , LHS is integer , RHS is not an integer when $x\gt2$,
If, $x=2\implies3^2-8=2^{2-2 }=1$, $x=2$ is a solution 
For all value of $x$ in negative integer , LHS is not an integer , RHS is an integer 
. $x=0$ is not a solution. 
$x=2$ is a solution 
