Question

If ${\log }_x{\log }_{18}\left(\sqrt{2}+\sqrt{8}\right)=\frac{1}{3}$. Then the value of $1000x$is equal to

Answer 1

${\log }_x{\log }_{18}\left(\sqrt{2}+\sqrt{8}\right)=\frac{1}{3}$

${\log }_x{\log }_{{\left(\sqrt{2}+\sqrt{8}\right)}^2}\left(\sqrt{2}+\sqrt{8}\right)=\frac{1}{3}$

${\log }_{{\left(\sqrt{2}+\sqrt{8}\right)}^2}\left(\sqrt{2}+\sqrt{8}\right)=x^{\frac{1}{3}}$

$\frac{1}{2}=x^{\frac{1}{3}}$

${\left(\frac{1}{2}\right)}^3=x$

$x=\frac{1}{8}$

$1000x=1000\times \frac{1}{8}$

$=125$