7.020. $\left(1+\frac{1}{2 x}\right) \log 3+\log 2=\log \left(27-3^{1 / x}\right)$.
Solution.

D: $\quad\left\{\begin{array}{l}x \neq 0, \\ 27-3^{1 / x}>0 .\end{array}\right.$
$\lg 3^{1+\frac{1}{2 x}}+\lg 2=\lg \left(27-3^{\frac{1}{x}}\right) \lg \left(2 \cdot 3^{1+\frac{1}{2 x}}\right)=\lg \left(27-3^{\frac{1}{x}}\right)$

$2 \cdot 3^{1+\frac{1}{2 x}}=27-3^{\frac{1}{x}}, 3^{\frac{1}{x}}+6 \cdot 3^{\frac{1}{2 x}}-27=0$.

This equation is quadratic in terms of $3^{\frac{1}{2 x}}$; we find $3^{\frac{1}{2 x}}=-9$, which is not valid, and $3^{\frac{1}{2 x}}=3$, from which $x=\frac{1}{2}$.

Answer: $\frac{1}{2}$.