What is the simplest way to tacke this ques?

gabriel: Legendary Member; Posts: 986; Joined: Wed Dec 20, 2006 11:07 am; Location: India; Thanked: 51 times; Followed by:1 members

Re: What is the simplest way to tacke this ques?

by gabriel » Fri Jul 13, 2007 4:49 am

Atult718 wrote:Hey guys,

How would you tackle this problem

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

remember (x*y)^n = x^n*y^n ...

so 10^n=(2*5)^n = 2^n*5^n ...

also 4^11 = (2^2)^11 = 2^22 ..

So ... we have 5^21*2^22= 2 * 2^n*5^n .. that is 5^21*2^21= 2^n*5^n .. comparing the power on both the sides of the equation we get n = 21 ...

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Atult718: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Wed Mar 14, 2007 1:17 am

by Atult718 » Fri Jul 13, 2007 6:06 am

thanks for the great explanation gabriel.

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