IPMAT Indore 2022 · Quantitative Ability · Question 16

Step 1 — simplify the left bound.

{\log}_{27}8=\frac{\log 8}{\log 27}=\frac{3\log 2}{3\log 3}={\log}_{3}2

So the left inequality is {\log}_{3}2\le{\log}_{3}x, i.e. x\ge 2 — and since the bound is not strict, x=2 is included.

Step 2 — simplify the right bound. Use \frac{1}{{\log}_{2}3}={\log}_{3}2:

{9}^{\frac{1}{{\log}_{2}3}}={9}^{{\log}_{3}2}={\left({3}^{2}\right)}^{{\log}_{3}2}={\left({3}^{{\log}_{3}2}\right)}^{2}={2}^{2}=4

So the right inequality is {\log}_{3}x<4, i.e. x<{3}^{4}=81 — strict, so 81 is excluded.

Step 3 — combine.

2\le x<81 \Rightarrow x\in [2, 81)

(The domain condition x>0 is already implied.)

Answer: Option 1 — [2, 81)

Shortcut. Two standard simplifications end this before any inequality work:

{\log}_{27}8={\log}_{{3}^{3}}{2}^{3}={\log}_{3}2 — matching exponents on base and argument cancel.

\frac{1}{{\log}_{2}3}={\log}_{3}2, so {9}^{{\log}_{3}2}={2}^{2}=4.

Then read the bracket types straight off the inequality signs: \le gives a closed left end, < an open right end — so [2,81).

All four options have the same shape, differing only in endpoints and brackets. That is a signal to be careful about \le versus < rather than to redo the algebra.