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 address this issue, we first analyze the two provided inequalities: \(5^{n-1} < 3^{n+1}\) and \(3^{n+1} < 2^{n+m}\). The objective is to determine the smallest integer value for \(m\) that satisfies the second inequality for all values of \(n\) that satisfy the first inequality.

Step 1: Simplify the first inequality

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

Rewriting this yields: \(\left(\frac{5}{3}\right)^{n-1} < 9\). This is equivalent to \(\frac{5}{3}^{n-1} < 9\). Given that \(\frac{5}{3} < 2\), this inequality is valid for all natural numbers \(n\), thus imposing no restriction on \(n\).

Step 2: Analyze the second inequality

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

Rewriting this expression gives: \(3 \cdot 3^n < 2^n \cdot 2^m\). Dividing both sides by \(3^n\) results in: \(3 < \left(\frac{2}{3}\right)^n \cdot 2^m\).

Step 3: Determine the smallest \(m\)

As \(n\) increases, the term \(\left(\frac{2}{3}\right)^n\) approaches 0. Consequently, \(2^m\) must be sufficiently large to ensure that the inequality \(3 < \left(\frac{2}{3}\right)^n \cdot 2^m\) remains true for all values of \(n\).

Step 4: Estimate suitable \(m\)

We will now test values to find the minimum \(m\) that satisfies the inequality:

• Test \(m = 3\): \(\left(\frac{2}{3}\right)^n \cdot 8\). This value decreases below 3 as \(n\) increases.
• Test \(m = 4\): \(\left(\frac{2}{3}\right)^n \cdot 16\). This also fails for larger values of \(n\).
• Test \(m = 5\): \(\left(\frac{2}{3}\right)^n \cdot 32\). This value exceeds 3 for all \(n\).

Final Answer: The minimum integer value for \(m\) is 5.