7.037. $\log _{6}\left(3^{x^{2}}+1\right)-\log _{6}\left(3^{2-x^{2}}+9\right)=\log _{6} 2-1$.
## Solution.

From the condition

$$
\begin{aligned}
& \log _{6}\left(3^{x^{2}}+1\right)-\log _{6}\left(3^{2-x^{2}}+9\right)=\log _{6} 2-\log _{6} 6, \log _{6} \frac{3^{x^{2}}+1}{3^{2-x^{2}}+9}=\log _{6} \frac{2}{6} \\
& \frac{3^{x^{2}}+1}{9 \cdot 3^{-x^{2}}+9}=\frac{2}{6}, 3^{2 x^{2}}-2 \cdot 3^{x^{2}}-3=0 .
\end{aligned}
$$

Solving this equation as a quadratic in terms of $3^{x^{2}}$, we get $3^{x^{2}}=-1$ (not valid) or $3^{x^{2}}=3$, from which $x^{2}=1, x_{1,2}= \pm 1$.

Answer: $-1 ; 1$.