help with a problem

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by truplayer256 » Mon Jul 27, 2009 7:13 am
5^21*4^11= 2*10^N

5^21*(2^2)^11=2*10^N. Remember that (x^m)^n=x^(m*n).

5^21*2^(2*11)=2*10^N

5^21*2^22=2*10^N

Divide both sides by 2.

5^21*2^21=10^N

(10)^(21)=(10)^(N). Remember that (m)^(x)*(n)^(x)=(m*n)^(x).

N=21

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by ghacker » Mon Jul 27, 2009 1:26 pm
what is the the value of n?

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

The important thing here is 10 can be written as 2*5

The L.H.S = R.H.S of the equation so if you look carefully you can see that we can make tens using the 5's and the 4's ( 4=2*2)

The limiting factor here is the number of 5's so the maximum power of 10 we can make is 21 because we have only 21 5's but 22 2's(4^11 =2^22)

Hence n = 21