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# (R^c)(R^d)(R^e) = R^(-12). If R>0, and c, d, and e are each different

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Math Expert
Joined: 02 Sep 2009
Posts: 62290
(R^c)(R^d)(R^e) = R^(-12). If R>0, and c, d, and e are each different  [#permalink]

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05 Nov 2017, 00:21
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72% (01:01) correct 28% (01:01) wrong based on 87 sessions

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$$(R^c)(R^d)(R^e) = R^{(-12)}$$. If R > 0, and c, d, and e are each different negative integers, what is the smallest that c could be?

A. -12
B. -10
C. -9
D. -6
E. -1

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Math Expert
Joined: 02 Aug 2009
Posts: 8298
Re: (R^c)(R^d)(R^e) = R^(-12). If R>0, and c, d, and e are each different  [#permalink]

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05 Nov 2017, 00:36
Bunuel wrote:
$$(R^c)(R^d)(R^e) = R^{(-12)}$$. If R > 0, and c, d, and e are each different negative integers, what is the smallest that c could be?

A. -12
B. -10
C. -9
D. -6
E. -1

hi..

$$(R^c)(R^d)(R^e) = R^{(-12)}.............R^{c+d+e}=R^{(-12)}$$..
equating the powers
c+d+e=-12....
All three are different and negative integers
so for c to be smallest, d and e have to be MAX possible..
so let d = -1 and e=-2..
so $$c+d+e=-12.......c-1-2=-12.....c=-12+3=-9$$

C
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Re: (R^c)(R^d)(R^e) = R^(-12). If R>0, and c, d, and e are each different   [#permalink] 05 Nov 2017, 00:36
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# (R^c)(R^d)(R^e) = R^(-12). If R>0, and c, d, and e are each different

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