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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]

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New post 22 Oct 2003, 18:57
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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
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New post 23 Oct 2003, 06:22
(-2)^(2m) = 4^m
and
2^(9-m) = 4^((9-m)/2)

Therefore,
m = (9-m)/2
2m = 9 - m
m = 3
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New post 23 Oct 2003, 15:13
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
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Re: if m is an integer such that (-2)^2m=2^(9-m) then m=?  [#permalink]

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New post 01 Sep 2013, 14:42
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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
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Re: if m is an integer such that (-2)^2m=2^(9-m) then m=?  [#permalink]

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New post 02 Sep 2013, 02:21
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


Yes, your reasoning is correct.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-m-is-an-integer-such-that-2-2m-2-9-m-then-m-144787.html
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Re: If m is an integer such that (-2)^2m=2^(9-m) then m=?  [#permalink]

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Re: If m is an integer such that (-2)^2m=2^(9-m) then m=? &nbs [#permalink] 02 Aug 2018, 21:14
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