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# If m is an integer such that (-2)^2m=2^(9-m) then m=?

Author Message
Intern
Joined: 07 Aug 2003
Posts: 18
Location: USA
If m is an integer such that (-2)^2m=2^(9-m) then m=?  [#permalink]

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22 Oct 2003, 17:57
1
2
00:00

Difficulty:

5% (low)

Question Stats:

92% (00:39) correct 8% (01:12) wrong based on 146 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. 5
E. 6

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-m-is-an-integer-such-that-2-2m-2-9-m-then-m-144787.html
Manager
Joined: 26 Aug 2003
Posts: 229
Location: United States

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23 Oct 2003, 05:22
(-2)^(2m) = 4^m
and
2^(9-m) = 4^((9-m)/2)

Therefore,
m = (9-m)/2
2m = 9 - m
m = 3
Manager
Joined: 11 Mar 2003
Posts: 54
Location: Chicago

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23 Oct 2003, 14:13
1
Hi Preyshi,

While posting a question, can you psot the question without answer? So that people will have chance to work on the problem?

Of course, it would be great if you can post the answer later after 1 or 2 people have tried the question. This way everybody can get the confirmation on answer.

Thanks
Intern
Joined: 18 Aug 2013
Posts: 15
Re: if m is an integer such that (-2)^2m=2^(9-m) then m=?  [#permalink]

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01 Sep 2013, 13:42
2
I know this is an old question, but I had a quick question on a potential variation to this problem. The only reason we were able to solve this question is because (-2)^2m, where m is an integer, is raised to an even power. Hence, (-2)^even = (2)^even, and we can get similar bases, and set 2m = 9m - 2.

What if (-2) was raised to an odd number times m? Let's say the equation was:

(-2)^3m = 2^(9-m)

Here, we cannot equal bases unless m = even integer. If we attempted to solve it, assuming m = even integer, we would get 3m = 9 - m; m = 9/4, which is not an even integer. So this problem has no solution.

Is this reasoning correct? Can we assume m = even integer (and hence, the bases are equal) and solve for m? If the equation leads to m = even integer, then would that be the solution, if not, then there is no solution?

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 52284
Re: if m is an integer such that (-2)^2m=2^(9-m) then m=?  [#permalink]

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02 Sep 2013, 01:21
grant1377 wrote:
I know this is an old question, but I had a quick question on a potential variation to this problem. The only reason we were able to solve this question is because (-2)^2m, where m is an integer, is raised to an even power. Hence, (-2)^even = (2)^even, and we can get similar bases, and set 2m = 9m - 2.

What if (-2) was raised to an odd number times m? Let's say the equation was:

(-2)^3m = 2^(9-m)

Here, we cannot equal bases unless m = even integer. If we attempted to solve it, assuming m = even integer, we would get 3m = 9 - m; m = 9/4, which is not an even integer. So this problem has no solution.

Is this reasoning correct? Can we assume m = even integer (and hence, the bases are equal) and solve for m? If the equation leads to m = even integer, then would that be the solution, if not, then there is no solution?

Thanks

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-m-is-an-integer-such-that-2-2m-2-9-m-then-m-144787.html
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 9426
Re: If m is an integer such that (-2)^2m=2^(9-m) then m=?  [#permalink]

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02 Aug 2018, 20:14
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If m is an integer such that (-2)^2m=2^(9-m) then m=? &nbs [#permalink] 02 Aug 2018, 20:14
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