CAT 2023 Slot 2 · Quant · Question 5

Deal with the quotient of logarithms first, since it collapses completely.

By the change-of-base rule, \frac{{\log}_{x}\left( 25 \right)}{{\log}_{x}\left( 0.008 \right)}={\log}_{0.008}\left( 25 \right).

Now 0.008=\frac{8}{1000}=\frac{1}{125}={5}^{-3}, and 25={5}^{2}.

So {\log}_{0.008}\left( 25 \right)={\log}_{{5}^{-3}}\left({5}^{2}\right)=\frac{2}{-3}=-\frac{2}{3}.

For the first term, \sqrt{3}={3}^{\frac{1}{2}}, so {\log}_{\sqrt{3}}\left( x \right)=\frac{{\log}_{3}x}{\frac{1}{2}}=2{\log}_{3}x.

The given equation is therefore 2{\log}_{3}x-\frac{2}{3}=\frac{16}{3}.

Adding \frac{2}{3} to both sides, 2{\log}_{3}x=\frac{18}{3}=6.

So {\log}_{3}x=3, which means x=27.

The quantity wanted is {\log}_{3}\left( 3{x}^{2}\right)={\log}_{3}3+{\log}_{3}{x}^{2}=1+2{\log}_{3}x.

Substituting, that is 1+2\times 3=7.

Shortcut. The target {\log}_{3}\left( 3{x}^{2}\right) equals 1+2{\log}_{3}x, and 2{\log}_{3}x is precisely the first term of the given equation. So the answer is just 1+\left( \frac{16}{3}+\frac{2}{3}\right)=1+6=7 — x itself is never needed.