7.076. $9^{\sqrt{x-5}}-27=6 \cdot 3^{\sqrt{x-5}}$.

7.076. $9^{\sqrt{x-5}}-27=6 \cdot 3^{\sqrt{x-5}}$.
Solution.

Domain of definition: $x-5 \geq 0, x \geq 5$.

$$
3^{2 \sqrt{x-5}}-6 \cdot 3^{\sqrt{x-5}}-27=0
$$

We solve the equation as a quadratic equation in terms of $3^{\sqrt{x-5}}$. We have $3^{\sqrt{x-5}}=-3$. (not suitable) or $3^{\sqrt{x-5}}=9$, from which $\sqrt{x-5}=2$, or $x-5=4$. Then $x=9$.

Answer: 9.