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What's the easiest way to solve this?

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What's the easiest way to solve this? [#permalink]

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New post 27 Mar 2010, 13:59
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If 5^21 * 4^11 = 2 * 10^n what is the value of n?

a) 11
b) 21
c) 22
d) 23
e) 32
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Re: What's the easiest way to solve this? [#permalink]

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New post 27 Mar 2010, 14:07
Ans b.
rewrite the question statement as - 5^21 * 2^21 * 2
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Re: What's the easiest way to solve this? [#permalink]

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New post 27 Mar 2010, 22:18
changhiskhan wrote:
If 5^21 * 4^11 = 2 * 10^n what is the value of n?
a) 11
b) 21
c) 22
d) 23
e) 32


5^21 * 2^(2*11) = 2*10^n
5^21 * 2^22 = 2*10^n
10^21 * 2 = 2*10^n (Multiplying 2^21 with 5^21)
so n = 21 hence B
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Re: What's the easiest way to solve this? [#permalink]

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New post 29 Mar 2010, 11:04
changhiskhan wrote:
If 5^21 * 4^11 = 2 * 10^n what is the value of n?

a) 11
b) 21
c) 22
d) 23
e) 32


Rewriting some of the terms:
\(4^{11} = (2*2)^{11} = 2^{22}
2*10^n = 2*(2*5)^n = 2*2^n*5^n= 2^{(n+1)}\)if equal the five to five and two with two
\(5^{21} = 5^n\)and \(2^{22} = 2^{(n+1)}\)

So n must be 21
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Re: What's the easiest way to solve this? [#permalink]

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New post 29 Mar 2010, 21:56
\(5^{21}*4^{11} = 2*10^n\)

Taking a "5" and a "4" out of their exponent sets lets me get to "2 x (2)(5)" which is starting to look like what I need:
\(5*5^{20}*4*4^{10} = 2*10^n\)
\(2*2*5*5^{20}*4^{10} = 2*10^n\)
\(2*10*5^{20}*4^{10} = 2*10^n\)

With what's left, there are two 2s in every 4. This gives me twenty 2s and twenty 5s - letting me make twenty 10s through multiplication:

\(2*10*10^{20} = 2*10^n\)

Reduce and solve for n:
\(2*10^{21} = 2*10^n\)
\(n = 21\) (b)
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Re: What's the easiest way to solve this? [#permalink]

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New post 29 Mar 2010, 22:00
Another way:

The number of "5"s on the left side has to equal that on the right side: \(5^{21} = 5^n\) --> n = 21
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Re: What's the easiest way to solve this?   [#permalink] 29 Mar 2010, 22:00
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