7.171. $5^{1+\log _{4} x}+5^{\log _{0.25} x-1}=\frac{26}{5}$.
Solution.

Domain of definition: $x>0$.

Let's switch to base 4. We have $5 \cdot 5^{\log _{4} x}+\frac{1}{5 \cdot 5^{\log _{4} x}}-\frac{26}{5}=0 \Leftrightarrow$ $\Leftrightarrow 25 \cdot\left(5^{\log _{4} x}\right)^{2}-26 \cdot 5^{\log _{4} x}+1=0 \Rightarrow\left(5^{\log _{4} x}\right)_{1}=5^{-2},\left(5^{\log _{4} x}\right)_{2}=5^{0}$, from which $\left(\log _{4} x\right)_{1}=-2,\left(\log _{4} x\right)_{2}=0$. Therefore, $x_{1}=\frac{1}{16}, x_{2}=1$.

Answer: $\frac{1}{16} ; 1$.