Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 166 / 1 / b

Solution

Put

Λ = n lo g 2 - m lo g 3

3^{m} \leq 2^{n - 1}

3^{m} \geq 2^{n + 1}

, the claimed inequality follows immediately after increasing the effective constant. We may therefore assume

2^{n - 1} < 3^{m} < 2^{n + 1},

which implies

m ≍ n

2

and

3

, gives

∣Λ∣ \geq n^{- C_{0}}

for an effective absolute constant

C_{0}

. If

∣Λ∣ > 1

, the desired conclusion is again immediate. Otherwise, the mean value theorem applied to the exponential function on

[- 1, 1]

gives

∣ e^{u} - 1∣ \geq c ∣ u ∣

. Hence

∣ 2^{n} - 3^{m} ∣ = 3^{m} ∣ e^{Λ} - 1∣ \geq c 3^{m} ∣Λ∣ \geq c 2^{n - 1} n^{- C_{0}} .

Absorbing the fixed factor into a larger exponent proves

∣ 2^{n} - 3^{m} ∣ \geq 2^{n} / n^{C}

Solved by gpt-5.6-sol high.

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