CAT 2023 Slot 3 QA Q7 Logarithms & Exponents

Question 7 of the Quantitative Ability (QA) section, CAT 2023 Slot 3, with CAT's official answer and a worked solution.

Let \(n\) be any natural number such that \(5^{n-1} \lt 3^{n+1}\). Then, the least integer value of \(m\) that satisfies \(3^{n+1} \lt 2^{n+m}\) for each such \(n\), is

TITA: type in the answer. There are no options and no negative marking.

Answer: 5 Logarithms & Exponents TITA

Solution

Check \(5^{n-1}<3^{n+1}\): n = 1 to 5 work (\(625<729\) at n = 5), n = 6 fails (\(3125>2187\)) and the gap only widens.
Need \(3^{n+1}<2^{n+m}\) for each n = 1..5. The tightest case is n = 5: \(729 < 2^{5+m}\) needs \(2^{5+m}\ge1024\), so \(m\ge5\).
Other cases: n = 3 (81 < 128) and n = 4 (243 < 256) need only m = 4. So least m = 5.