Question

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

Answer 1

${\log }_x{\log }_{18}\sqrt{2}+\sqrt{8}=\frac{-1}{2}$

${\log }_{18}\sqrt{2}+\sqrt{2^3}=x^{-1/2}$

${\log }_{18}\sqrt{2}+\sqrt{(2{)}^3}=x^{-1/2}$

${\log }_{18}\sqrt{2}\left(\right(\sqrt{2}{)}^2+1)=x^{-1/2}$

${\log }_{18}\sqrt{2}(2+1)=x^{-1/2}$

${\log }_{18}\sqrt{2}\times 3=x^{-1/2}$

${\log }_{18}\sqrt{18}=x^{-1/2}$

$\frac{1}{2}{\log }_{18}18=x^{-1/2}$

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

$(\frac{1}{2}{)}^{-2}=(x)$

$x=4$