Question

Solve for $x$

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

Answer 1

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

$\Rightarrow 5^{1+{\log }_{4}x}+5^{-{\log }_{4}x-1}=\frac{26}{5}$

Let $t=5^{1+{\log }_{4}x}$

$\Rightarrow t+\frac{1}{t}=\frac{26}{5}$

$\Rightarrow \frac{t^2+1}{t}=\frac{26}{5}$

$\Rightarrow 5t^2-26t+5=0$

$\Rightarrow 5t^2-25t-t+5=0$

$\Rightarrow 5t(t-5)-1(t-5)=0$

$\Rightarrow (t-5)(5t-1)=0$

$\Rightarrow t=5,\frac{1}{5}$

$\Rightarrow 5^{1+{\log }_{4}x}=5,5^{1+{\log }_{4}x}=\frac{1}{5}=5^{-1}$

$\Rightarrow 1+{\log }_{4}x=1,1+{\log }_{4}x=-1$

$\Rightarrow {\log }_{4}x=0,{\log }_{4}x=-2$

$\Rightarrow x=4^0=1,4^{-2}=\frac{1}{16}$

$\therefore x=1,\frac{1}{16}$