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Exponent problem

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Intern
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Exponent problem [#permalink] New post 06 Feb 2007, 16:21
If 5^21X4^11=2X10^n, what is n?

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21
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The answer is 21, but why? Step by step please. Thank you.
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 [#permalink] New post 06 Feb 2007, 16:54
I'm no math expert, but here is how I did it.

The first thing I did was rename 4^21 as 2^22. This is because 4=2^2 and you multiply the exponents.

The second step was to divide both sides by 2, to get 5^21 * 2^21 = 10^n

When I saw this I realized that what you're saying is you have 21 groups of 5 and 21 groups of 2. Since 5 * 2=10, you have 21 groups of 10 or 10^21.

I don't know if there is an exponent rule for the last step, but that's how I solved it.
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 [#permalink] New post 06 Feb 2007, 17:47
Thanks for the explanation
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 [#permalink] New post 07 Feb 2007, 00:59
If 5^21X4^11=2X10^n, what is n?

11
21
22
23
32

5^21 * 2^22 -1 = 2^n*5*n

thus 2^21 = 2^n , n = 21
  [#permalink] 07 Feb 2007, 00:59
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