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# If 5^21 X 4^11 = 2 X 10^n, what is the valve of n?

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Manager
Joined: 05 Aug 2008
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Schools: McCombs Class of 2012
If 5^21 X 4^11 = 2 X 10^n, what is the valve of n? [#permalink]

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17 Dec 2008, 15:53
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If $$5^2^1 X 4^1^1 = 2 X 10^n$$, what is the valve of n?

(A) 11
(B) 21
(C) 22
(D) 23
(E) 32

E is correct

anyone please give an approach to how they solved this problem (problem attached as image).

thanks.

Ops -- sorry wrong place!
Attachments

Q1.JPG [ 14.24 KiB | Viewed 784 times ]

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Intern
Joined: 17 Dec 2008
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17 Dec 2008, 17:30
I got stuck on this one as well.

The way it was explained is:

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

1. Notice that 4^21 can be factored as 2 x 2^21
2. Bases with the same exponent can be multiplied together, keeping the same exponent (5x2 = 10 ... add the exponent back 10^21)
3. That leaves you with 2 (that you factored out from step 1) and 10^21 AKA

2 x 10^21

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Retired Moderator
Joined: 05 Jul 2006
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17 Dec 2008, 23:28
5^21 * 4^11 = 2* 10^n

5^21*2^21*2 = 2*10^n,

(5*2)^21 = 10^n

10^21 = 10^n

n = 21

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Re: Exponents   [#permalink] 17 Dec 2008, 23:28
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# If 5^21 X 4^11 = 2 X 10^n, what is the valve of n?

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