7.176. $2^{\log _{5} x^{2}}-2^{1+\log _{5} x}+2^{\log _{5} x-1}-1=0$.

7.176. $2^{\log _{5} x^{2}}-2^{1+\log _{5} x}+2^{\log _{5} x-1}-1=0$.
Solution.

Domain of definition: $x>0$.

Rewrite the equation as $2^{2 \log _{5} x}-2 \cdot 2^{\log _{5} x}+\frac{2^{\log _{5} x}}{2}-1=0 \Leftrightarrow$ $\Leftrightarrow 2 \cdot 2^{2 \log _{5} x}-3 \cdot 2^{\log _{5} x}-2=0$. Solving this equation as a quadratic equation in terms of $2^{\log _{5} x}$, we find $2^{\log _{5} x}=-\frac{1}{2}$ (not suitable) or $2^{\log _{5} x}=2$, from which $\log _{5} x=1, x=5$.

Answer: 5.