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Can this sum be solved with this method?

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by faraz_jeddah » Tue Jul 16, 2013 5:27 am
[email protected] wrote:(-2)^2m=2^9-m then m =?
Can we solve it this way if not then why?
-2^2m= 4^m And
2^9-m is equal to 4^7-m, now 4^m=4^7-m therefore m= 7-m, which is equal to 2m=7 therefore m 7/2
The text in red is your mistake.

take m = 6

2^(9-m) = 2^(9-6) = 2^3 = 8
4^(7-m) = 4^(7-6) = 4^1 = 4
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by Brent@GMATPrepNow » Tue Jul 16, 2013 6:06 am
[email protected] wrote: If m is an integer, and (-2)^(2m) = 2^(9-m) then m = ?
NOTE: I added some brackets in order to make the exponents clear. Also, in the original version of this question, we are told that m is an integer. So, I added that as well. Okay, now onto the solution.

First, it's important to recognize that, if m is an integer, then (-2)^(2m) = 2^(2m)
One way to prove this is to first rewrite -2 as (-1)(2)
So, we get: (-2)^(2m) = [(-1)(2)]^(2m)
= [(-1)^(2m)][(2)^(2m)] by applying the Power of a Product rule
IMPORTANT: If m is an integer, then 2m is an even power, which means (-1)^(2m) = 1
So, we get: = 1[(2)^(2m)]
= (2)^(2m)

So, now that we know that (-2)^(2m) = 2^(2m), we can take the original equation, (-2)^(2m) = 2^(9-m), and rewrite it as:
2^(2m) = 2^(9-m)
Since the bases are the same, we know that 2m = 9 - m
Solve to get m = 3

Cheers,
Brent
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by Matt@VeritasPrep » Tue Jul 16, 2013 9:56 am
We'd do it like this:

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

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

4^m = 2^(9-m)

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

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

2m = 9 - m

3m = 9

m = 3
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