Bunuel wrote:

If \((\frac{5}{4})^{-n} < 16^{-1}\). What is the least integer value of n?

Well my approach is as follows-

Let us rearrange the equation

\((\frac{4}{5})^{n} <\frac{1}{16}\)

or \((\frac{4}{5})^{n} <(\frac{1}{2})^{4}\) (since \(1^{any power}\) = 1)

now \(\frac{4}{5} =0.8\)which is greater than\(\frac{1}{2}\) i.e 0.5

Now \((\frac{4}{5})^{3}\) =0.512 is slightly greater than \(\frac{1}{2}\) i.e 0.5

or \((\frac{4}{5})^{3*4} >(\frac{1}{2})^{4}\) (we have to take the RHS of equality to 1/16) and ( \(\frac{4}{5} = 0.8\) and \(0.8^3=0.512\))

or \((\frac{4}{5})^{12} >(\frac{1}{2})^{4}\)

Therefore to make it less we need to multiply one more 4/5

i.e \((\frac{4}{5})^{13} < (\frac{1}{2})^{4}\) (because when we multiply a number smaller than 1 exponentially, the value becomes less)

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