Question

If $lo{g}_{\sqrt{2}}\sqrt{x}+lo{g}_{2}x+lo{g}_4\left(x^2\right)+lo{g}_8\left(x^3\right)+lo{g}_{16}\left(x^4\right)=40$ then $x$ is equal to

Answer 1

${\log }_{\sqrt{2}}\sqrt{x}+{\log }_{2}x+{\log }_4{x}^2+{\log }_8{x}^3+{\log }_{16}{x}^4=40$
$\frac{{\log }_2\sqrt{x}}{{\log }_2\sqrt{2}}+{\log }_{2}x+\frac{{\log }_2{x}^2}{{\log }_{2}4}+\frac{{\log }_2{x}^3}{{\log }_{2}8}+\frac{{\log }_2{x}^4}{{\log }_{2}16}=40$
$\frac{{\log }_2\sqrt{x}}{{\log }_2{2}^{1/2}}+{\log }_{2}x+\frac{{\log }_2{x}^2}{{\log }_2{2}^2}+\frac{{\log }_2{x}^3}{{\log }_2{2}^3}+\frac{{\log }_2{x}^4}{{\log }_2{2}^4}=40$
$2{\log }_2\sqrt{x}+{\log }_{2}x+\frac{1}{2}{\log }_2{x}^2+\frac{1}{3}{\log }_2{x}^3+\frac{1}{4}{\log }_2{x}^4=40$
$\frac{2}{2}{\log }_{2}x+\frac{2}{2}{\log }_{2}x+\frac{2}{2}{\log }_{2}x+\frac{3}{3}{\log }_{2}x+\frac{4}{4}{\log }_{2}x=40$
${\log }_{2}x+{\log }_{2}x+{\log }_{2}x+{\log }_{2}x+{\log }_{2}x=40$
$5{\log }_{2}x=40$
${\log }_{2}x=8$
$2^{{\log }_{2}x}=2^8$
$x=2^8$
$x=256$