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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 Updated on: 28 May 2013, 12:41
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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

Originally posted by fozzzy on 27 Dec 2012, 06:46.
Last edited by walker on 28 May 2013, 12:41, edited 3 times in total.
edited the question.
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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 27 Dec 2012, 06:55
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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 13 Sep 2014, 06:18
Hi Bunuel, hi other members,

first of all thanks a lot everybody who is and has been posting and answering questions here which is a great support to my preparation. I learned a lot from you guys.

Question on this one, I got answer D (4) and am not sure why exactly I'm wrong - could somebody comment? I calculated like this:

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

<=> 4^m = 4^8-m

<=> m = 8-m

<=> m = 4


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 13 Sep 2014, 08:55
1
christian1904 wrote:
Hi Bunuel, hi other members,

first of all thanks a lot everybody who is and has been posting and answering questions here which is a great support to my preparation. I learned a lot from you guys.

Question on this one, I got answer D (4) and am not sure why exactly I'm wrong - could somebody comment? I calculated like this:

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

<=> 4^m = 4^8-m

<=> m = 8-m

<=> m = 4


Thanks!

You made one small mistake \(2^{9-m} \ne 4^{8-m}\).

\(2^{9-m} = \frac{2^9}{2^m} = \frac{2^{2 * 4.5}}{2^m} = \frac{4^{4.5}}{2^m}\)

So, what you could have done was \(4^m = \frac{4^{4.5}}{2^m}\)

=> \(2^m = \frac{4^{4.5}}{4^m}\)

=> \(2^m = 4^{4.5 - m}\)

=> \(2^m = 2^{2*(4.5 - m)}\)

=> \(m = 2*(4.5 - m)\)

=> \(m = 9 - 2m\)

=> \(m = 3\)

But Bunuel's way was much faster.

Hope that helps.
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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 13 Feb 2015, 13:34
does it matter that one base is negative and one base is positive? just asking for future similar problems because i did the problem without noticing that there was a negative sign in front of the first 2. i got the problem right still by just doing 2m=9-m, but im sure if i noticed he negative sign i wouldve freaked out a little.
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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 16 Feb 2015, 03:32
btan219 wrote:
does it matter that one base is negative and one base is positive? just asking for future similar problems because i did the problem without noticing that there was a negative sign in front of the first 2. i got the problem right still by just doing 2m=9-m, but im sure if i noticed he negative sign i wouldve freaked out a little.


Two things:
1. When the bases are the same you can equate their powers.
2. Negative numbers in even powers are positive, thus (-2)^even = 2^even.
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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 16 Feb 2015, 23:20
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.

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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=?   [#permalink] 08 Aug 2019, 13:46
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