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 Post subject: Arithmatic
PostPosted: Sun Dec 04, 2011 11:05 am 
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Joined: Sun Dec 04, 2011 10:53 am
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If (1/5)to the power of m times (1/4) to the power of 18 = 1/2(10) to the power of 35. What is m? I attahced a screen shot

This question is from the GMART PREP exam.


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 Post subject: Re: Arithmatic
PostPosted: Mon Dec 05, 2011 1:38 pm 
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Joined: Thu Dec 01, 2011 6:36 am
Posts: 6
Tough one, but nice... here what my thoughts are:
you know that 10 = 2x5

so (1/10)^35 = (1/2)^35 * (1/5)^35

Now we know that (2^2)^x = 2^(2x), visa versa, 2^Y=(2^2)^(y/2)

Back to our equation:
(1/2)^35 = (1/(2^2))^(35/2) = (1/4)^(35/2)

Now you have the 1/2 * (1 / 4)^(35/2) * (1/5)^35... I hope this writing format is clear enough...

I can also express (1/2) = (1/4)^(1/2)... using the same thought...

Multiplying (1/4)^(1/2) with (1/4)^(35/2) = (1/4)^(1/2+35/2) = (1/4)^(36/2) = (1/4)^(18)

Back to the equation:
(1/5)^m * (1/4)^18 = (1/5)^35 * (1/4)^18

Cancel out both (1/4)^18, and I hope I am correct, but it will be (1/5) ^m = (1/5) ^35
or in other words, m=35...

The trick I am going by, is trying to match both sides parameters, to make sure I am dealing with the m power only...


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 Post subject: Re: Arithmatic
PostPosted: Mon Dec 05, 2011 8:41 pm 
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Joined: Thu Feb 12, 2009 6:32 pm
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(1/5)to the power of m times (1/4) to the power of 18 = 1/2(10) to the power of 35.

(1/5)^m is the same as 5^(-m)
(1/4)^18 = 4^(-18) ==> convert the 4 to 2^2, and this becomes 2^(-36)

On the right, we have
2^(-1) * 10^(-35)
the 10 becomes (2*5) so 10^(-35) becomes 2^(-35) * 5^(-35)
Our right side is now:
2^(-1) * 2^(-35) * 5^(-35)
Combining the 2's, we have
2^(-36) * 5^(-35)

Our whole equation is now:
2^(-36) * 5^(-m) = 2^(-36) * 5^(-35)
The 2^(-36) cancels from each side, leaving us
5^(-m) = 5^(-35)
which means m = 35.

Veritas Help


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