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by papgust » Tue Nov 17, 2009 10:07 pm
5^21 * 4^11 = 2 * 10^n

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

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

Equate either 2 or 5,

2^22 = 2^(n+1)
22=n+1
n=21

5^21 = 5^n
n=21

Hence, B
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by chipbmk » Tue Nov 17, 2009 10:18 pm
papgust wrote:5^21 * 4^11 = 2 * 10^n

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

5^21 * 2^22 = 2^(n+1) * 5^n
Thanks again! You are awesome!

I had gotten it as far as I bolded above, but I did not know how to compare the two different things, now I know you can split them out separately.

Thanks!
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by Testluv » Tue Nov 17, 2009 10:19 pm
chipbmk wrote:If 5^21 * 4^11 = 2 * 10^n , what is n?

a. 11
b. 21
c. 22
d. 23
e. 32

OA: B

Please give a detailed explanation of how to solve.

Thanks!
In these problems, we try to make the bases equivalent. After that, somerules you will have to use commonly are:

(a^b)^c = a^b*c; and

a*c * b*c = (a*b)^c (when you have the same exponents, you can "group" the bases together and raise them to the common exponent)

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

5^21 * (2^2)^11 = RHS

5^21 * 2^22 = RHS

5^21 * 2^21 * 2^1 = RHS

(5*2)^21 * 2^1 = RHS

2 * 10^21 = 2*10^n

21 = n
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