Question

Solve the following equation:
$2(\log 2-1)+\log (5^{\sqrt{x}}+1)=\log (5^{1-\sqrt{x}}+5)$

Answer 1

$2(\log 2-1)+\log (5^{\sqrt{x}}+1)=\log (5^{1-\sqrt{x}}+5)$

$2\log \left(\frac{2}{e}\right)+\log (5^{\sqrt{x}}+1)=\log (5^{1-\sqrt{x}}+5)$

$takeanti\log andlet{5}^{\sqrt{x}}=z$

$\frac{4}{e^2}.(z+1)=\frac{5}{z}+5$

$z(4z+4)=(5+5z)e^2$

$4z^2+4z-5z{e}^2-5e^2=0$

$4z^2+z(4-5z^2)-5e^2=0$

$z=\frac{-(4-5e^2)\pm \sqrt{{(4-5e^2)}^2+4\times 4\times 5e^2}}{2\times 4}$

$z=\frac{-(4-5e^2)\pm (4+5e^2)}{8}$

$z=\frac{-(4-e^2)+(4+5e^2)}{8}andz=\frac{-(4-5e^2)-(4+5e^2)}{8}$

$z=\frac{10e^2}{8},z=-1$

$thus,5^{\sqrt{x}}=\frac{10e^2}{8}$

$take\log$

$\sqrt{x}=\frac{(\log 10+2-\log 8)}{\log 5}$

$=>x={({\log }_5\frac{10e^2}{8})}^2$