rk

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rk

by R.K » Mon May 28, 2012 8:47 am
If 2^n=128,then (2^(n-1))(5^(n-1))=?

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by Anurag@Gurome » Mon May 28, 2012 8:50 am
R.K wrote:If 2^n=128,then (2^(n-1))(5^(n-1))=?
2^n = 128 ---> n = 7

Hence, (2^(n - 1))*(5^(n - 1)) = (2*5)^(n - 1) = 10^(7 - 1) = 10^6 = 1,000,000
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