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# 1/ , then m=? TIA

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1/ , then m=? TIA [#permalink]

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18 Apr 2007, 22:55
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

[(1/5)^m]*[(1/4)^18]= 1/[2(10^35)], then m=?

TIA

Last edited by ayl989 on 19 Apr 2007, 13:20, edited 2 times in total.
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18 Apr 2007, 23:41
Hi,

I am not clear on what does "to the" mean

if X to the Y means x/y then the answer should be 35/19

on solving we will have

19*M/20 = 35/20 or M = 35/19....

regards,

Amardeep
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19 Apr 2007, 00:11
Try reposting your question. If you mean 2 to the power of y, then type 2^y.
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19 Apr 2007, 11:07
sorry for the confusion, I've edited the question.
thanks again.
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19 Apr 2007, 11:49
ayl989 wrote:
[(1/5)^m]*[(1/4)^19]= 1/[2(10^35)], then m=?

TIA

[(1/5)^m]*[(1/4)^19]= 1/[2(10^35)],

5^m * 4^19 =2*(5*2)^35

5^m * (2*2)^19 =2*(5*2)^35

5^m * 2^38 =2^36*5^35

5^m * 2^2 =5^35

4 * (5^m ) = 5^35

4 = 5^(35-m)
Dunno how to solve further
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19 Apr 2007, 12:03
Hi Ay1989

What are the options?? are u still missing something?

regards,

Amardeep
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19 Apr 2007, 13:21
whoops, edited the question and changed 19 to 18

17
18
34
35
36

so from your work, I see how the 2^36 cancels out and then m =35.
I always have problems with these, but it seems like radicals are all about factoring out the original problem and trying to isolate into similar expressions.
thanks again.
19 Apr 2007, 13:21
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