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# 1/ , then m=? TIA

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Intern
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1/ , then m=? TIA [#permalink]  18 Apr 2007, 21:55
[(1/5)^m]*[(1/4)^18]= 1/[2(10^35)], then m=?

TIA

Last edited by ayl989 on 19 Apr 2007, 12:20, edited 2 times in total.
Director
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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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Try reposting your question. If you mean 2 to the power of y, then type 2^y.
Intern
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sorry for the confusion, I've edited the question.
thanks again.
VP
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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
Director
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Hi Ay1989

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

regards,

Amardeep
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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.
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