Question

If ${\log }_{x}a$, $a^{x/2}$, ${\log }_{b}x$ are om $G.P$. Then $x$ is equal to

• A. ${\log }_a\left({\log }_{b}a\right)$
• B. $\frac{\log \left(\log a\right)-\log \left(\log b\right)}{\log \left(\log a\right)}$
• C. ${\log }_b\left({\log }_{a}b\right)$
• D. $Noneofthese$

Answer 1

The correct option is A ${\log }_a\left({\log }_{b}a\right)$

Given that,

${\log }_{x}a{,}^{x/2},{\log }_{b}x$ are in G.P.

So,

${\left(a^{x/2}\right)}^2=\left({\log }_{x}a\right)\left({\log }_{b}x\right)$

$a^x=\frac{\left(\log a\right)}{\log x}\times \frac{\log x}{\log b}[\because {\log }_{a}x=\frac{\log x}{\log a}]$

$a^x=\frac{\left(\log a\right)}{\log b}$

$a^x={\log }_{b}a$

$x{\log }_{a}a={\log }_a\left({\log }_{b}a\right)$

$x={\log }_a\left({\log }_{b}a\right)$

Hence, this is the answer.