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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # (5/4)^-n < 16^-1 What is the least integer value of n?  Question banks Downloads My Bookmarks Reviews Important topics
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Senior Manager Joined: 20 May 2014
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(5/4)^-n < 16^-1 What is the least integer value of n? [#permalink] 00:00

Question Stats: 0% (00:00) correct 100% (02:39) wrong based on 3 sessions
If $$(\frac{5}{4})^{-n} < 16^{-1}$$. What is the least integer value of n?

[Reveal] Spoiler:
13

Kudos for correct solution. Director Joined: 03 Sep 2017
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Re: (5/4)^-n < 16^-1 What is the least integer value of n? [#permalink]
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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. Director Joined: 20 Apr 2016
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Re: (5/4)^-n < 16^-1 What is the least integer value of n? [#permalink]
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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]
IlCreatore wrote:
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

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

Please explain why is 64/5 Re: (5/4)^-n < 16^-1 What is the least integer value of n?   [#permalink] 20 Dec 2018, 05:21
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