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

Updated On: Sep 17, 2024
  • 5
  • 6
  • 7
  • None of Above
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The Correct Option is A

Solution and Explanation

Given a natural integer n, it can be shown that \(5^{ 𝑛 βˆ’ 1} < 3 ^{𝑛 + 1}\).
 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.

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