7.6 $\quad 2^{x^{2}-1}-3^{x^{2}}=3^{x^{2}-1}-2^{x^{2}+2}$.
7.6 Let's group the powers with base 2 on the left side and the powers with base 3 on the right side of the equation, and factor each part. We have

$2^{x^{2}-1}\left(1+2^{3}\right)=3^{x^{2}-1}(1+3)$, or $2^{x^{2}-1} \cdot 3^{2}=3^{x^{2}-1} \cdot 2^{2}$.

Dividing both sides of the equation by $3^{x^{2}-1} \cdot 2^{2} \neq 0$, we get $\frac{2^{x^{2}-1} \cdot 3^{2}}{3^{x^{2}-1} \cdot 2^{2}}=1$, or $\frac{2^{x^{2}-3}}{3^{x^{2}-3}}=1$, or $\left(\frac{2}{3}\right)^{x^{2}-3}=\left(\frac{2}{3}\right)^{0}$.

From this, $x^{2}-3=0$, i.e., $x_{1,2}= \pm \sqrt{3}$.

Answer: $x_{1,2}= \pm \sqrt{3}$.