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(5/4)^-n < 16^-1 What is the least integer value of n?
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01 Oct 2017, 22:19
01 Oct 2017, 22:19
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32% (02:01) correct
67% (01:57) wrong based on 40 sessions
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If \((\frac{5}{4})^{-n} < 16^{-1}\). What is the least integer value of n?
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Re: (5/4)^-n < 16^-1 What is the least integer value of n?
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02 Oct 2017, 09:23
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.
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.
Re: (5/4)^-n < 16^-1 What is the least integer value of n?
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02 Oct 2017, 11:27
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)
