CAT 2023 Slot 2 QA Question 6: For some positive real number $x$, if $\log_{\sqrt{3}}(x) + \dfrac{\log_{x}(25)}{\log_{x}(0.008)} = \dfrac{16}{3}$, then the value of $\log_{3}(3x^{2})$ is

Question

For some positive real number $x$, if $\log_{\sqrt{3}}(x) + \dfrac{\log_{x}(25)}{\log_{x}(0.008)} = \dfrac{16}{3}$, then the value of $\log_{3}(3x^{2})$ is

Accepted Answer

The second term is independent of the base $x$ because it is a ratio of logarithms with the same base.

$\dfrac{\log_{x}(25)}{\log_{x}(0.008)} = \dfrac{\log(25)}{\log(0.008)}$
$25 = 5^{2}$
$0.008 = \dfrac{1}{125} = 5^{-3}$
$\dfrac{\log(25)}{\log(0.008)} = \dfrac{2}{-3} = - \dfrac{2}{3}$

Substitute

For some positive real number xxx, if log⁡3(x)+log⁡x(25)log⁡x(0.008)=163\log_{\sqrt{3}}(x) + \dfrac{\log_{x}(25)}{\log_{x}(0.008)} = \dfrac{16}{3}log3​​(x)+logx​(0.008)logx​(25)​=316​, then the value of log⁡3(3x2)\log_{3}(3x^{2})log3​(3x2) is

For some positive real number $x$ , if $\log_{\sqrt{3}}(x) + \dfrac{\log_{x}(25)}{\log_{x}(0.008)} = \dfrac{16}{3}$ , then the value of $\log_{3}(3x^{2})$ is