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If m is an integer such that (-2)^2m = 2^{9 - m}, then m=

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Joined: 02 Sep 2009
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If m is an integer such that (-2)^2m = 2^{9 - m}, then m=  [#permalink]

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14 Mar 2014, 03:23
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If m is an integer such that $$(-2)^{2m} = 2^{9 - m}$$, then m=

(A) 1
(B) 2
(C) 3
(D) 4
(E) 6

Problem Solving
Question: 166
Category: Algebra First -degree equations; Exponents
Page: 84
Difficulty: 600

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Re: If m is an integer such that (-2)^m = 2^{9 - m}, then m=  [#permalink]

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14 Mar 2014, 03:23
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SOLUTION

If m is an integer such that $$(-2)^{2m} = 2^{9 - m}$$, then m=

(A) 1
(B) 2
(C) 3
(D) 4
(E) 6

First of all, since m is an integer, then 2m=even, and therefore $$(-2)^{2m}=2^{2m}$$.

Sol, we have that $$2^{2m} = 2^{9 - m}$$ --> $$2m=9-m$$ --> $$m=3$$.

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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m=  [#permalink]

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14 Mar 2014, 17:50
1
So, I choose C. Here is my reasoning:

-2^2m = 2^9-2
from this - (-2)^2m is the same as 2^2m, since the power is even number, so we have
2m = 9-m
3m = 9
m = 3
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Joined: 20 Oct 2015
Posts: 36
Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m=  [#permalink]

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05 Jul 2016, 05:19
[quote="Bunuel"]The Official Guide For GMAT® Quantitative Review, 2ND Edition

If m is an integer such that $$(-2)^{2m} = 2^{9 - m}$$, then m=

(A) 1
(B) 2
(C) 3
(D) 4
(E) 6

2^{9-m} = 2^9 / 2^m
(-2)^2m= 2^ 2m

2^2m * 2^m = 2^9

2m + m = 9

m=3
C
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Joined: 30 Sep 2016
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m=  [#permalink]

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03 Dec 2016, 05:50
What if the question was (-2)^3m = 2^9-m?
Is this possible?
Math Expert
Joined: 02 Sep 2009
Posts: 60687
Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m=  [#permalink]

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03 Dec 2016, 05:55
1
Nasahtahir wrote:
What if the question was (-2)^3m = 2^9-m?
Is this possible?

No real m satisfies that equation.
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m=  [#permalink]

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04 Sep 2019, 18:44
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If m is an integer such that $$(-2)^{2m} = 2^{9 - m}$$, then m=

(A) 1
(B) 2
(C) 3
(D) 4
(E) 6

Recall that when a negative number is raised to an even power, the result is positive. Since 2m is even, (-2)^(2m) = 2^(2m), and we have:

2^(2m) = 2^(9 - m)

2m = 9 - m

3m = 9

m = 3

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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m=  [#permalink]

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05 Sep 2019, 04:30
since 2m is even, then 9-m must be even
we can exclude all even numbers : 2,4,6, we left with 1 and 3
2*1 = 2
9-1 = 8
so one is not the solution
2*3 = 6
9-3 = 6
so m = 3 .
Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m=   [#permalink] 05 Sep 2019, 04:30
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