IPMAT Indore 2026 · Quantitative Ability · Question 43

Step 1 — pull the 2 inside the first logarithm.

2{\log}_{3}\left( 3n \right)={\log}_{3}{\left( 3n \right)}^{2}={\log}_{3}9{n}^{2}

Step 2 — combine the two logarithms.

{\log}_{3}9{n}^{2}-{\log}_{3}\left({n}^{2}+1 \right)={\log}_{3}\left( \frac{9{n}^{2}}{{n}^{2}+1}\right)

Step 3 — take n large. Divide numerator and denominator by {n}^{2}:

\frac{9{n}^{2}}{{n}^{2}+1}=\frac{9}{1+\frac{1}{{n}^{2}}}\to 9 \text{as}n\to \infty

Step 4 — evaluate.

{\log}_{3}9=2

Answer: Option 1 — 2

Shortcut. For “sufficiently large n”, the +1 is noise: treat {n}^{2}+1 as {n}^{2} straight away, and the expression is {\log}_{3}\frac{9{n}^{2}}{{n}^{2}}={\log}_{3}9. One step.