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

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If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   10 Jul 2020, 08:00
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If m is an integer such that (−2)2m=29−m, then m=

(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   10 Jul 2020, 08:00
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   12 Jul 2020, 23:02
Since base is same so,
2m = 9 - m
2m+m = 9
3m = 9
m = 3

Just wondering about minus sign with 2 in question stem. We have to ignore it?
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   10 Aug 2020, 07:44
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Farina wrote:
Since base is same so,
2m = 9 - m
2m+m = 9
3m = 9
m = 3

Just wondering about minus sign with 2 in question stem. We have to ignore it?



The base is not the same.
The base of the first one is -2 for the LHS and 2 for the RHS.
But since the LHS is raised to an even power the negative doesn't matter and we can treat it as 2 instead of -2

Why? B/c

(-2)^2m = ((-1)(2))^2m = (-1)^2m * (2)^2m = ((-1)^2)^m * (2)^2m = 1^m * (2)^2m = 1 * (2)^2m = (2)^2m
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink] New post   10 Aug 2020, 16:44
chacinluis wrote:
Farina wrote:
Since base is same so,
2m = 9 - m
2m+m = 9
3m = 9
m = 3

Just wondering about minus sign with 2 in question stem. We have to ignore it?



The base is not the same.
The base of the first one is -2 for the LHS and 2 for the RHS.
But since the LHS is raised to an even power the negative doesn't matter and we can treat it as 2 instead of -2

Why? B/c

(-2)^2m = ((-1)(2))^2m = (-1)^2m * (2)^2m = ((-1)^2)^m * (2)^2m = 1^m * (2)^2m = 1 * (2)^2m = (2)^2m


Thank you very much for clarification
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Re: If m is an integer such that (-2)^2m = 2^{9 - m}, then m= [#permalink]
10 Aug 2020, 16:44
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