IPMAT Indore 2026 · Quantitative Ability · Question 32

Step 1 — move everything to one base. Work in base 2 and write t={\log}_{2}3.

{\log}_{18}24=\frac{{\log}_{2}24}{{\log}_{2}18}

Since 24={2}^{3}\cdot 3 and 18=2\cdot{3}^{2}:

p=\frac{3+t}{1+2t}

Step 2 — solve for t.

3+t=p+2pt \Rightarrow t\left( 1-2p \right)=p-3 \Rightarrow t=\frac{p-3}{1-2p}

Step 3 — express the target the same way. Since 108={2}^{2}\cdot{3}^{3} and 96={2}^{5}\cdot 3:

{\log}_{96}108=\frac{2+3t}{5+t}

Step 4 — substitute.

\text{numerator}=2+\frac{3\left( p-3 \right)}{1-2p}=\frac{2-4p+3p-9}{1-2p}=\frac{-p-7}{1-2p}

\text{denominator}=5+\frac{p-3}{1-2p}=\frac{5-10p+p-3}{1-2p}=\frac{2-9p}{1-2p}

{\log}_{96}108=\frac{-p-7}{2-9p}=\frac{p+7}{9p-2}

Answer: Option 2 — \frac{p+7}{9p-2}

Shortcut. Reduce every number to {2}^{x}{3}^{y} and let t={\log}_{2}3 carry the whole question — then it is ordinary algebra. If you are short of time, take p={\log}_{18}24\approx 1.0995 on the calculator-free estimate {\log}_{2}3\approx 1.585 and test the options numerically; only one lands near {\log}_{96}108\approx 1.0256.