Bunuel wrote:
If \((\frac{5}{4})^{-n} < 16^{-1}\). What is the least integer value of n?
A. 12
B. 13
C. 14
D. 15
E. 16
\((\frac{5}{4})^{-n} < 16^{-1}\)
=> A = \((\frac{4}{5})^n*2^4 < 1\)
For n = 0, A =16; n = 1, A = 16*0.8.
So, as the value of n increases, A decreases and as n is increased by 1, A decreases by 80 %. We need to look for the value of n for which A >1 and A*80% less than 1.
As the power of 2 is 4, I was looking to test multiple of 4 for n while solving. So, I decided to first test with n = 12; from the list of option. As it will bit simplify A.
For n = 12, A = \((\frac{4}{5})^{12}*2^4\) = \((\frac{4^3}{5^3})^4*2^4\) = \((\frac{64}{125})^4*2^4\)
= \((\frac{128}{125})^4\)
A will be slightly greater than 1. But, definitely 80% of A will be just greater than 0.8. So, if n=13, value of A will be around 0.8 and it is the least integer value for which the inequality satisfies.
Hence, B. I might be a bit lucky.
as the question was definitely not an easy one to crack.