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Exponents Problem- Please help

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by DeepakR » Wed Oct 29, 2008 8:57 am
I think answer is m=35 !

In the question in RHS (right hand side) you have 1/[(2*10)^35] which can be written as 1/{[2^35]*[(5*2)^35]} which will get cancelled with LHS value of 1/4^35=1/(2^70). Thus you will have something like (1/5)^m=(1/5)^35 hence m=35

- Deepak
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Thanks, but I still don't get it :-)

by Posey MBA11 » Wed Oct 29, 2008 9:41 am
Hi Deepak,

The answer is 35 but can you PLEASE break this down to more manageable parts?

Thanks!
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by jsl » Thu Oct 30, 2008 2:50 am
(1/5)^m (1/4)^35 = 1/{2(10)^35)}

equals...
(1^m)(5^-m)(1^35)(4^-35) = (2^-1)(10^-35)

equals...
(5^-m)(1^35)(4^-35) = (2^-1)(10^-35)

equals...
(5^-m)(1^35)(4^-35) = (2^-1)(2^-35)(5^-35)

equals...
(5^-m)(4^-35) = (2^-36)(5^-35)

equals...
(5^-m)(2^-70) = (2^-36)(5^-35)

equals...
(5^-m) = (2^-34)(5^-35)

...that's as far as I got.... am I going on the right track? If not, where did I go wrong?
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Re: Exponents Problem- Please help

by yezz » Thu Oct 30, 2008 7:04 am
If (1/5)^m (1/4)^35 = 1/{2(10)^35)} , then m=

17
18
34
35
36

1 / 2(10)^35 = 2^-36*5^-35 thus m = 35
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by sam77 » Fri Oct 31, 2008 4:41 pm
jsl wrote:(1/5)^m (1/4)^35 = 1/{2(10)^35)}

equals...
(1^m)(5^-m)(1^35)(4^-35) = (2^-1)(10^-35)

equals...
(5^-m)(1^35)(4^-35) = (2^-1)(10^-35)

equals...
(5^-m)(1^35)(4^-35) = (2^-1)(2^-35)(5^-35)

equals...
(5^-m)(4^-35) = (2^-36)(5^-35)

equals...
(5^-m)(2^-70) = (2^-36)(5^-35)

equals...
(5^-m) = (2^-34)(5^-35)

...that's as far as I got.... am I going on the right track? If not, where did I go wrong?
Typo in question. It should be 1/[(2*10)^35] on RHS
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