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

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Joined: 01 Nov 2017
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The smallest integer n for which 4^n > 17^19 holds, is closest to  [#permalink]

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Updated on: 23 Feb 2019, 19:30
2
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25% (medium)

Question Stats:

74% (01:26) correct 26% (01:35) wrong based on 102 sessions

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

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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.

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

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Updated on: 18 Apr 2019, 18:36
$$2^{2n}>17^{19} = (2^{4}+1)^{19}$$

$$2^{2n}>2^{76}$$

n must be more than 38.

E

Originally posted by jfranciscocuencag on 23 Feb 2019, 14:43.
Last edited by jfranciscocuencag on 18 Apr 2019, 18:36, edited 2 times in total.
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The smallest integer n for which 4^n > 17^19 holds, is closest to  [#permalink]

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

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03 May 2019, 22:07
raghavrf wrote:
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

$$4^n$$ > [m]17^{19}[/m
4^n>(16)^19+1^19
4^n>4^38+1^19
n>=38
n=39
IMOE
Re: The smallest integer n for which 4^n > 17^19 holds, is closest to   [#permalink] 03 May 2019, 22:07
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# The smallest integer n for which 4^n > 17^19 holds, is closest to

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