Question

solve $5^{1+lo{g}_{4}x}+5^{lo{g}_{25}x-1}=\frac{26}{5}$

Answer 1

If $lo{g}_{4}x=t$ then
$lo{g}_{25}x=lo{g}_4^{-1}x=-lo{g}_{4}x=-t$
$\therefore 5^{1+t}=5^{-(1+t)}=\frac{26}{5}$
Put $5^{(1+t)}=z\therefore z+\frac{1}{z}=\frac{26}{5}$
or $5z^2-26z+1=0\therefore z=5,1/5or=5^{-1}$
$\therefore 5^{1+t}=5^{1}or{5}^{-1}\therefore 1+t=1or-1$
$Hence,t=0or-2\therefore lo{g}_{4}x=0or-2\therefore x=4^{\circ }or{4}^{-2}orx=1,1/16$