Question

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

• A. 5
• B. 6
• C. 7
• D. None of Above

Answer 1

The Correct Option is A

To solve this problem, we begin by analyzing the two inequalities given: \(5^{n-1} < 3^{n+1}\) and \(3^{n+1} < 2^{n+m}\). Our task is to find the smallest integer value of \(m\) that satisfies the second inequality for all \(n\) that satisfy the first inequality.

Step 1: Simplify the first inequality 

\(5^{n-1} < 3^{n+1}\)

Rewriting: \(\left(\frac{5}{3}\right)^{n-1} < 9\)

This implies \(\frac{5}{3}^{n-1} < 9\). Since \(\frac{5}{3} < 2\), the inequality holds for all natural numbers \(n\). So, no restriction is needed for \(n\).

Step 2: Analyze the second inequality

\(3^{n+1} < 2^{n+m}\)

Rewriting: \(3 \cdot 3^n < 2^n \cdot 2^m\)

Divide both sides by \(3^n\): \(3 < \left(\frac{2}{3}\right)^n \cdot 2^m\)

Step 3: Determine the smallest \(m\)

As \(n\) increases, \(\left(\frac{2}{3}\right)^n \to 0\). Therefore, \(2^m\) must be large enough to ensure the inequality \(3 < \left(\frac{2}{3}\right)^n \cdot 2^m\) holds for all \(n\).

Step 4: Estimate suitable \(m\)

Let us check the minimum value that ensures the inequality holds:

• Try \(m = 3\): \(\left(\frac{2}{3}\right)^n \cdot 8\) → becomes smaller than 3 as \(n\) increases.
• Try \(m = 4\): \(\left(\frac{2}{3}\right)^n \cdot 16\) → still fails for large \(n\).
• Try \(m = 5\): \(\left(\frac{2}{3}\right)^n \cdot 32\) > 3 for all \(n\).

Final Answer: The least integer value of \(m\) is 5.