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14 Mar 2014, 03:23
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
96% (01:03) correct
4%
(01:37)
wrong
based on 2724
sessions
History
Date
Time
Result
Not Attempted Yet
If m is an integer such that \((-2)^{2m} = 2^{9 - m}\), then m=
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
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 \((-2)^{2m}=2^{2m}\).
So, we'd have:
\(2^{2m} = 2^{9 - m}\);
\(2m=9-m\);
\(m=3\).
Answer: C.
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 \((-2)^{2m}=2^{2m}\).
So, we'd have:
\(2^{2m} = 2^{9 - m}\);
\(2m=9-m\);
\(m=3\).
Answer: C.
General Discussion
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
-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
