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Help with Quant Q (greater explanation)

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Help with Quant Q (greater explanation) [#permalink]

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New post 29 Jan 2009, 11:42
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can someone help explain in greater detail Step 2 and 3 in the below?

(1/5)^m*(1/4)^18 = 1/2(10)^35
Step 1) (1/5)^m*(1/2)^36 = 1/2(10)^35 {since 2^2=4}
Step 2)(1/5)^m*(1/2)^m*(1/2)^(36-m) = 1/2*1/(10)^35
Step 3)(1/10)^m*(1/2)^(36-m) = 1/2*(1/10)^35
equate indices from both sides. i.e. equate index of (1/2) and (1/10).
left hand side =1/2*(1/5^35)*(1/2)^35 hence, m=35

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Re: Help with Quant Q (greater explanation) [#permalink]

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New post 29 Jan 2009, 12:48
jaycurtis wrote:
Can someone help explain in greater detail Step 2 and 3 in the below?

(1/5)^m*(1/4)^18 = 1/2(10)^35
Step 1) (1/5)^m*(1/2)^36 = 1/2(10)^35 {since 2^2=4}
Step 2)(1/5)^m*(1/2)^m*(1/2)^(36-m) = 1/2*1/(10)^35
Step 3)(1/10)^m*(1/2)^(36-m) = 1/2*(1/10)^35
equate indices from both sides. i.e. equate index of (1/2) and (1/10).
left hand side =1/2*(1/5^35)*(1/2)^35 hence, m=35


That's an unconventional solution - nothing wrong with it, but I'm not sure many people would think to introduce the fraction(1/2)^m at step 2, especially since there isn't any reason to. You can do as follows, which I'd personally find simpler, using prime factorizations to ensure we have the same base numbers:

\(\left(\frac{1}{5} \right)^m \times \left( \frac{1}{4} \right)^{18} = \frac{1}{2 \times 10^{35}\)

\(\left(\frac{1}{5^m} \right) \times \left( \frac{1}{(2^2)^{18}} \right) = \frac{1}{2 \times (2 \times 5)^{35}\)

\(\frac{1}{5^m \times 2^{36} } = \frac{1}{2 \times 2^{35} \times 5^{35}\)

\(\frac{1}{5^m \times 2^{36} } = \frac{1}{2^{36} \times 5^{35}\)

\(m = 35\)
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Re: Help with Quant Q (greater explanation) [#permalink]

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New post 30 Jan 2009, 04:24
Everything is clear!

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Re: Help with Quant Q (greater explanation) [#permalink]

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New post 30 Jan 2009, 09:30
Thanks that makes much more sense!!!

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Re: Help with Quant Q (greater explanation)   [#permalink] 30 Jan 2009, 09:30
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