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

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Bunuel
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If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   14 Mar 2014, 03:23
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A
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C
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5% (low)

Question Stats:

96% (01:02) correct 4% (01:36) wrong based on 2010 sessions
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If m is an integer such that \((-2)^{2m} = 2^{9 - m}\), then m=

(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
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C
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Bunuel
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   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 \(2^{2m} = 2^{9 - m}\) --> \(2m=9-m\) --> \(m=3\).

Answer: C.
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ershovici
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   14 Mar 2014, 17:50
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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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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   16 Feb 2015, 23:20
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Hi All,

While this question might look a little "scary", the answer choices are small numbers and we're just asked to find the one that completes the given equation, so we can use "brute force" on this question.

We're looking for the value of M that makes (-2)^(2M) = 2^(9-M)

We can work through the answers in order (although you might realize that the first couple are much TOO small, given the exponents in this prompt).

IF....
M = 1
(-2)^2 = 4
2^8 = 256
These are NOT equal

M = 2
(-2)^4 = 16
2^7 = 128
These are NOT equal

M = 3
(-2)^6 = 64
2^6 = 64
These ARE equal, so this MUST be the answer.

Final Answer:
Show Spoiler
C


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Rich
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   05 Jul 2016, 05:19
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[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 :-D
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   03 Dec 2016, 05:50
What if the question was (-2)^3m = 2^9-m?
Is this possible?
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Bunuel
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   03 Dec 2016, 05:55
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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] New post   04 Sep 2019, 18:44
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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

Answer: C
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   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 .
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If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   11 Apr 2021, 20:38
Bunuel wrote:
Nasahtahir wrote:
What if the question was (-2)^3m = 2^9-m?
Is this possible?


No real m satisfies that equation.


Not trying to beat a dead horse, but in general, what happens when we have a negative base? How would we deal with it if the goal is to solve for an unknown?
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If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   18 May 2021, 02:16
(-2)^2m=2^(9-m)
=> (-1*2)^2m= 2^(9-m)
=> (-1)^2m +2^2m=2^(9-m)
=>2^2m=2^(9-m)
=> 2m= 9-m
=> m=3
* whatever m is odd or even the power of -1 will always even number as multiplied by 2 and resulting in +1
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   14 Oct 2021, 20:15
Bunuel wrote:
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 \(2^{2m} = 2^{9 - m}\) --> \(2m=9-m\) --> \(m=3\).

Answer: C.


I don't understand how we can just set these two exponents equal to eachother. -2 and 2 are not the same bases. This violates the rules. Can you explain?
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Bunuel
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   15 Oct 2021, 01:15
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kaylaquijas wrote:
Bunuel wrote:
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 \(2^{2m} = 2^{9 - m}\) --> \(2m=9-m\) --> \(m=3\).

Answer: C.


I don't understand how we can just set these two exponents equal to eachother. -2 and 2 are not the same bases. This violates the rules. Can you explain?



This is explained in the highlighted part, no?
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   15 Oct 2021, 05:36
Bunuel wrote:
kaylaquijas wrote:
Bunuel wrote:
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 \(2^{2m} = 2^{9 - m}\) --> \(2m=9-m\) --> \(m=3\).

Answer: C.


I don't understand how we can just set these two exponents equal to eachother. -2 and 2 are not the same bases. This violates the rules. Can you explain?



This is explained in the highlighted part, no?


I guess I'm just having a hard time understanding it. Thank you.
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If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   04 Oct 2023, 04:29
Bunuel wrote:
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 \(2^{2m} = 2^{9 - m}\) --> \(2m=9-m\) --> \(m=3\).

Answer: C.


Bunuel

What if 'm' was a proper or improper fraction, would it still work?
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If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink]
04 Oct 2023, 04:29
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