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(5/4)^-n < 16^-1 What is the least integer value of n?

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(5/4)^-n < 16^-1 What is the least integer value of n? [#permalink] New post 01 Oct 2017, 22:19
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If \((\frac{5}{4})^{-n} < 16^{-1}\). What is the least integer value of n?


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13


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Re: (5/4)^-n < 16^-1 What is the least integer value of n? [#permalink] New post 02 Oct 2017, 09:23
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My guess:

Given \((\frac{5}{4})^{-n}<16^{-1}\), I try to have similar numbers on the two sides of the inequality. Thus, I multiply both sides by \(\frac{64}{5}\), in order to get

\((\frac{4}{5})^{n}\frac{64}{5}<\frac{4}{5}\).

Then, I divide both sides by \((\frac{4}{5})^{n}\) to get \(\frac{64}{5}<\frac{4}{5}^{1-n}\).

Finally, to get an easier intuition, I rewrite the expression as \(\frac{64}{5}<\frac{5}{4}^{n-1}\).

Now, \(\frac{64}{5} = 13^{-}\), while \(\frac{5}{4} = 1.25\). Thus, we know we need to raise \(\frac{5}{4}\) to a number around 11 or 12, since we need to reach less than 13 using a number greater than 1. Let's try.

If the exponent were 11, then n = 12 and we would get \(13^{-} > 11^{+}\). It is not enough.

If the exponent were 12, then n = 13 and we would get \(13^{-} < 14^{+}\). Here we are!

The minimal exponent that satisfies the inequality is 12, which means a minimal n = 13.
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Re: (5/4)^-n < 16^-1 What is the least integer value of n? [#permalink] New post 02 Oct 2017, 11:27
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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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Re: (5/4)^-n < 16^-1 What is the least integer value of n?   [#permalink] 02 Oct 2017, 11:27
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