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exponent problem

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by ikaplan » Fri Jul 15, 2011 3:47 pm
OK, here's my solution.

5^21 x 4^11 = 2 x 10^n
5^21 x (2^2)^11= 2 x (2 x 5)^n
5^21 x (2)^22= 2 x (2 x 5)^n
5^21 x (2)^21 x 2= 2 x (2 x 5)^n
(5 x 2)^21 x 2= 2 x (2 x 5)^n [cancel the 2s)

therefore,

n=21
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by MBA.Aspirant » Fri Jul 15, 2011 10:09 pm
hillscottc wrote:For the following question, 5^21 x 4^11 = 2 x 10^n, what is the value of n?
The answer is 21. How? Could someone provide an explanation how to figure this one?

Thanks.
prime factorize

5^21 x 4^11 = 2 x 10^n

5^21 * 2^11 * 2^11 = 2 * 2^n * 5^n

5^21 * 2^22 = 2^n+1 * 5^n

so either n+1 =22, n = 21

or n = 21 directly from the base 5
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