If ${\log _x}a,\;{a^{x/2}}$ and ${\log _b}x$ are in $G.P.$, then $x = $

  • A

    $ - \log ({\log _b}a)$

  • B

    $ - {\log _a}({\log _a}b)$

  • C

    ${\log _a}({\log _e}a) - {\log _a}({\log _e}b)$

  • D

    ${\log _a}({\log _e}b) - {\log _a}({\log _e}a)$

Similar Questions

If the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of a $G.P.$ are $a, b$ and $c,$ respectively. Prove that

$a^{q-r} b^{r-p} c^{p-q}=1$

Let $a_{n}$ be the $n^{\text {th }}$ term of a G.P. of positive terms.

If $\sum\limits_{n=1}^{100} a_{2 n+1}=200$ and $\sum\limits_{n=1}^{100} a_{2 n}=100,$ then $\sum\limits_{n=1}^{200} a_{n}$ is equal to 

  • [JEE MAIN 2020]

If ${(p + q)^{th}}$ term of a $G.P.$ be $m$ and ${(p - q)^{th}}$ term be $n$, then the ${p^{th}}$ term will be

The sum of the first three terms of a $G.P.$ is $S$ and their product is $27 .$ Then all such $S$ lie in 

  • [JEE MAIN 2020]

Let for $n =1,2, \ldots \ldots, 50, S _{ a }$ be the sum of the infinite geometric progression whose first term is $n ^{2}$ and whose common ratio is $\frac{1}{(n+1)^{2}}$. Then the value of $\frac{1}{26}+\sum\limits_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)$ is equal to

  • [JEE MAIN 2022]