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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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    Well, you can guess at one solution without much fuss. Can you show that there is only the one? – 2017-02-07
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    To be clear: there won't, in general, be a simple closed formula for the result. For example, if you replace the right hand with $2^{3-x}$ the unique solution is now $x \approx 2.0853813771459838801232593$, by numerical means. – 2017-02-07
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    Plotting both sides of the exponential form of the equation shows that there is an unique solution: http://www.wolframalpha.com/input/?i=3%5Ex-8%3D2%5E(2-x) – 2017-02-07

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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: $\{2\}$.

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$\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