7.122. $\sqrt[3]{27^{5 \sqrt{x}}}=3^{x(\sqrt{x}-4)}$.

7.122. $\sqrt[3]{27^{5 \sqrt{x}}}=3^{x(\sqrt{x}-4)}$.
Solution.

Domain of definition: $x \geq 0$.

We have $3^{5 \sqrt{x}}=3^{x(\sqrt{x}-4)} \Rightarrow 5 \sqrt{x}=x(\sqrt{x}-4) \quad \sqrt{x}=0, x_{1}=0, \quad$ or $(\sqrt{x})^{2}-4 \sqrt{x}-5=0$.

Solving this equation as a quadratic equation in terms of $\sqrt{x}$, we get $\sqrt{x}=-1, \varnothing$; or $\sqrt{x}=5, x=25$.

Answer: $0 ; 25$.