7.055. $\log _{3}\left(3^{x}-8\right)=2-x$.

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7.055. $\log _{3}\left(3^{x}-8\right)=2-x$.
Solution.

Domain of definition: $3^{x}-8>0$.

By the definition of logarithm, we have $3^{x}-8=3^{2-x}, 3^{x}-8=\frac{9}{3^{x}}$, $3^{2 x}-8 \cdot 3^{x}-9=0$, from which, solving this equation as a quadratic equation in terms of $3^{x}$, we find $3^{x}=-1, \varnothing$; or $3^{x}=9$, from which $x=2$.

Answer: 2.