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

Answer: (A)

5

Answer 2

Given a natural integer n, it can be shown that $5^{ 𝑛 − 1} < 3 ^{𝑛 + 1}$.
We can conclude from observation that the inequality is true for n = 1, 2, 3, 4, and 5.
Currently, we must determine the smallest integer value of m such that $3^{ n+1} <2^{ n+m}$ is satisfied.
In the case of n = 1, m's least integer value is 2.
In the case of n = 2, m's least integer value is 3.
In the case of n = 3, m's least integer value is 4.
In the case of n = 4, m's least integer value is 4.
In the case where n = 5, m's least integer value is 5.
Thus, 5 is the lowest integer value of m for which the equation holds true for all values of n.
The correct option is (A): 5.