driving me nutts

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driving me nutts

by cris » Thu Jan 31, 2008 9:30 am
1/5)^m (1/4)^18 = 1/2(10)^35

(1/5^m) (1/2^2)18=1/2(2*5)^35

(1/5^m) (1/2^36)=1/2*2^35*5^35

(1/5^m)=(1/5^35)

therefore m=35


Could someone please please please tell me step by step, how to go from here:
1/5^m) (1/2^36)=1/2*2^35*5^35

to here: (1/5^m)=(1/5^35)

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Re: driving me nutts

by Stuart@KaplanGMAT » Thu Jan 31, 2008 9:53 am
cris wrote:1/5)^m (1/4)^18 = 1/2(10)^35

(1/5^m) (1/2^2)18=1/2(2*5)^35

(1/5^m) (1/2^36)=1/2*2^35*5^35

(1/5^m)=(1/5^35)

therefore m=35


Could someone please please please tell me step by step, how to go from here:
1/5^m) (1/2^36)=1/2*2^35*5^35

to here: (1/5^m)=(1/5^35)
Unless the question was misquoted, the reason you're going crazy is because there is no way to get that solution.

(1/5)^m * (1/2)^36 = (1/2) * 2^35 * 5^35

(1/5)^m * (1/2)^36 = 2^34 * 5^35

(1/5)^m = (2^34 * 5^35)/(1/2)^36

(1/5)^m = (2^34 * 5^35) * 2^36

(1/5)^m = 2^70 * 5^35

5^(-m)/5^35 = 2^70

5^(-m-35) = 2^70

As you can see, there is no simple solution for m. If we backsolve with m=35, we get:

5^(-70) = 2^70

which clearly isn't true.


Does someone have the source of the question? Something must have been lost or transposed in the translation.
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by cris » Thu Jan 31, 2008 10:04 am
Thank god !! :)

Stuart, this question was posted by SALMA, Jan 29 2008 under the title "[b]Powers[/b]"....you can find it easly in the posts if you go a bit downwards....

Someone posted that solution and Slama said that is was ok....and I just couldnt see how...

Once again Stuart, thank you

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by sibbineni » Thu Jan 31, 2008 3:11 pm
I think the question is

(1/5)^m (1/4)^18 = 1/2(10)^35....

m=35 is the answer

Suggest me if iam wrong

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by Stuart@KaplanGMAT » Thu Jan 31, 2008 3:22 pm
sibbineni wrote:I think the question is

(1/5)^m (1/4)^18 = 1/2(10)^35....

m=35 is the answer

Suggest me if iam wrong
Is the right side supposed to be:

1/(2*(10^35))?

If the whole expression is in the denominator, then the question makes sense.

We now have:

(1/5)^m * (1/2)^36 = 1/(2 * 2^35 * 5^35)

(1/5)^m * (1/2)^36 = 1/(2^36 * 5^35)

(1/5)^m * (1/2)^36 = (1/2)^36 * (1/5)^35

Dividing both sides by (1/2)^36, we get:

(1/5)^m = (1/5)^35

and therefore m=35
Last edited by Stuart@KaplanGMAT on Thu Jan 31, 2008 3:27 pm, edited 1 time in total.
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by sibbineni » Thu Jan 31, 2008 3:25 pm
yes