CAT 2023 Slot 3 · Quant · Question 1

Each of the last two terms is a ratio of logarithms with the same base, which the change-of-base rule collapses.

So \frac{{\log}_{3}\left({2}^{x}-9 \right)}{{\log}_{3}4}={\log}_{4}\left({2}^{x}-9 \right).

Likewise the third term is {\log}_{4}\left({2}^{x}+\frac{17}{2}\right).

The first term also hides a logarithm: since {4}^{\frac{1}{2}}=2, \frac{1}{2}={\log}_{4}2.

So the three terms are {\log}_{4}2, {\log}_{4}\left({2}^{x}-9 \right) and {\log}_{4}\left({2}^{x}+\frac{17}{2}\right).

Logarithms in arithmetic progression mean their arguments are in geometric progression.

So {\left({2}^{x}-9 \right)}^{2}=2\times \left({2}^{x}+\frac{17}{2}\right)={2}^{x+1}+17.

Writing t={2}^{x}, this is {t}^{2}-18t+81=2t+17.

Rearranging, {t}^{2}-20t+64=0, which factors as \left( t-4 \right)\left( t-16 \right)=0.

If t=4 then {2}^{x}-9=-5, and the logarithm of a negative number is undefined — so this root is rejected.

So t=16, that is x=4.

The common difference is the second term minus the first: {\log}_{4}\left( 16-9 \right)-{\log}_{4}2={\log}_{4}7-{\log}_{4}2.

Combining, that is {\log}_{4}\frac{7}{2} — the third option.

Shortcut. Reading \frac{1}{2} as {\log}_{4}2 is the whole trick: all three terms become logarithms to base 4, so the arguments form a GP. The common difference is then the logarithm of the common ratio, \frac{7}{2}, read straight off.