7.072. $10^{\frac{2}{x}}+25^{\frac{1}{x}}=4.25 \cdot 50^{\frac{1}{x}}$.
Solution.

Domain of definition: $x \neq 0$.

Dividing both sides of the equation by $25^{\frac{1}{x}}$, we get $2^{\frac{2}{x}}-4.25\left(2^{\frac{1}{x}}\right)+1=0$, from which, solving the equation as a quadratic in terms of $2^{\frac{1}{x}}$, we obtain $\left(2^{\frac{1}{x}}\right)=\frac{1}{4}$, hence $\left(\frac{1}{x}\right)_{1}=-2, x_{1}=-\frac{1}{2}$, or $\left(2^{\frac{1}{x}}\right)_{2}=4$, hence $\left(\frac{1}{x}\right)_{2}=2, x_{2}=\frac{1}{2}$. Answer: $x_{1}=-\frac{1}{2} ; x_{2}=\frac{1}{2}$.