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# Exponents

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Intern
Joined: 25 Apr 2009
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25 Apr 2009, 14:58
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Please could someone help me, I have the following problem and would be grateful if someone could show me how to solve it:

(20)^n = 5^21 x 4^11

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Manager
Joined: 08 Feb 2009
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26 Apr 2009, 10:24
question doesn't seem right. Is "n" an integer ?

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Manager
Joined: 02 Mar 2009
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27 Apr 2009, 08:39
Yes..I dont know if the question is right:

(20)^n = 5^21 x 4^11

(2^2 x 5)^n = 5^21 x 2^22

Now equating the two:
2n = 22
n = 11

BUT

n=21 when we see the exponent of 5. Is something wrong?

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Re: Exponents   [#permalink] 27 Apr 2009, 08:39
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