logarithmic equation $\log_2(3^x-8)=2-x$

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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.

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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: $\{2\}$.