[tex]\displaystyle 5^{1+\log_4 x}+5^{-\log_4x-1} = \frac{26}{5}\\ \\
\text{Sea } z=5^{1+\log_4 x} \text{ entonces}: \\ \\
z+\frac{1}{z}=\frac{26}{5}\\ \\
5z^2-26z+5=0 \\ \\
(5z-1)(z-5)=0 \\ \\
z\in\left\{\frac{1}{5};5\right\}\\ \\
\text{Then: }\\ \\
5^{1+\log_4x}=5^{-1} \vee 5^{1+\log_4x}=5\\ \\
1+\log_4x=-1 \vee 1+\log_4x=1\\ \\
\log_4x =-2 \vee \log_4x=0 \\ \\
x=\frac{1}{16} \vee x=1\\ \\
\text{Answer: }x\in\left\{\frac{1}{16};1\right\}
[/tex]
