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The smallest integer n for which 4^n > 17^19 holds, is closest to

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The smallest integer n for which 4^n > 17^19 holds, is closest to  [#permalink]

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New post Updated on: 23 Feb 2019, 19:30
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The smallest integer n for which \(4^n\) > \(17^{19}\) holds, is closest to

(a) 31
(b) 33
(c) 35
(d) 37
(e) 39


Source: testbook.com

Originally posted by raghavrf on 23 Feb 2019, 05:09.
Last edited by u1983 on 23 Feb 2019, 19:30, edited 2 times in total.
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Re: The smallest integer n for which 4^n > 17^19 holds, is closest to  [#permalink]

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New post 23 Feb 2019, 09:49
since 17 is close to 16. \(17^{19}\) is approximately = \(16^{19}\) = \(4^{2(19)}\) = \(4^{38}\)
If \(4^n\) has to be greater than \(17^{19}\), n has to greater than 38, hence n = 39.

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Re: The smallest integer n for which 4^n > 17^19 holds, is closest to  [#permalink]

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New post 23 Feb 2019, 14:43
\(2^{2n}>17^{19} = (2^{4}+1)^{19}\)

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The smallest integer n for which 4^n > 17^19 holds, is closest to  [#permalink]

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New post 23 Feb 2019, 19:26
\(17^{19}\) > \(16^{19}\) or \(17^{19}\) > \(4^{38}\)
So for the inequality \(4^n\) > \(17^{19}\) to hold true, n must be GT 38
Hence n is closest to 39 (out of the given choices :-) )
Ans E
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The smallest integer n for which 4^n > 17^19 holds, is closest to   [#permalink] 23 Feb 2019, 19:26
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