aiq wrote:hello everybody.....

5^21 * 4^11 = 2 * 10^n, what is the value of n?
a)11
b)21
c)22
d)23
e)32

thanks a ton in advance....

A good solution above. There are a few ways to get to an answer here, and when you see these kinds of equations, with numbers and large exponents, one foolproof method is to break everything down into primes:

5^21 * 4^11 = 2 * 10^n
5^21 * (2^2)^11 = 2 * (2*5)^n
5^21 * 2^22 = 2 * 2^n * 5^n
5^21 * 2^22 = 2^(n+1) * 5^n

and since we have prime factorizations on both sides, the exponents must be equal on both sides - n = 21.

It's a good question, and illustrates how important it is to have a good command of exponent rules.