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Math Question

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One more question..

by kashewman » Sun Sep 10, 2006 11:00 am
Does anyone have a good formula for mixture problems? Thanks.
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POWER

by aim-wsc » Sun Sep 10, 2006 11:48 am
...err can you make some things clear?

like: 2(10)^35 did you mean [(2(10))^35]=20^35
or 2 X (10)^35

OK i presume its latter one.

Its a tricky one & i must confess i had to used my instincts

before i explain the answer, you must understand the following rule

given n>m
then
(a^m) X (b^n) = ((aXb)^m) X (b^(n-m))

eg. (5^4) X (3^6) = ((5X3)^4) X (3^2)
= (15 ^4) X 9

OK back to problem.
LHS:
(1/5)^m * (1/4)^18
this can be also wirtten as
:arrow: (1/5)^m * (1/2)^36

RHS:
1/[2 X 10^35]= 1/[2 X (2X5)^35)]
:arrow: =1/[(2^36) X (5^35)]


ANSWER:
therefore
m=35
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Re: One more question..

by beatthegmat » Sun Sep 10, 2006 7:34 pm
kashewman wrote:Does anyone have a good formula for mixture problems? Thanks.
Check out page 48 of my GMAT Flashcards: https://www.beatthegmat.com/viewtopic.php?t=32
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MAth

by abby_g » Sat Sep 30, 2006 2:06 am
The answer is : m = 35

RHS: 1/2(10)^35 = (1/5^35 )* (1/2^36)
= (1/5^35) * (1/4^18)

equate on both sides and woyla you get m = 35. Correct?
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