Question

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

• A. $8$
• B. $\frac{1}{8}$
• C. $\frac{1}{125}$
• D. $125$

Answer 1

The correct option is D $125$

${\log }_x{\log }_{18}(\sqrt{2}+\sqrt{8})=\frac{1}{3}$
${\log }_x{\log }_{18}(\sqrt{2}+2\sqrt{2})=\frac{1}{3}$
${\log }_x{\log }_{18}\left(3\sqrt{2}\right)=\frac{1}{3}$
${\log }_{18}\left(3\sqrt{2}\right)=x^{\frac{1}{3}}$
$\frac{\log \left(3\sqrt{2}\right)}{\log 18}=x^{\frac{1}{3}}$
$\frac{\log \left(3\sqrt{2}\right)}{\log (3\sqrt{2}{)}^2}=x^{\frac{1}{3}}$
$\frac{1}{2}=x^{\frac{1}{3}}$
$\Rightarrow x=\frac{1}{8}$
$\therefore 1000x=1000\times \frac{1}{8}=125$.