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

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5^21 * 4^11= 2* 10^n [#permalink] New post 25 Aug 2003, 11:39
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5^21 * 4^11= 2* 10^n


solve for n?
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 [#permalink] New post 25 Aug 2003, 11:57
Just to show work:

5^21 * 4^11 = 2* 10^n
5^21 * 2^22 = 2* 10^n
2(5^21 * 2^21) = 2*10^n
(5^21 * 2^21) = 10^n
10^21 = 10^n
n=21
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 [#permalink] New post 25 Aug 2003, 12:04
theres a riskier way to solve this... just by noticing that 10=2*5, so if n is to be an integer, n must be 21
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 [#permalink] New post 29 Aug 2003, 06:11
Just to show work:

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

How do you go from: 5^21 * 2^22 = 2* 10^n

to: 2(5^21 * 2^21) = 2*10^n

Please explain.

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 [#permalink] New post 29 Aug 2003, 06:26
All i did was separate 2^22 into 2(2^21).
As a rule, 2(2^x) = 2^(x+1).
Which leads me to another interesting problem I saw..I'll put it in a new thread.
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 [#permalink] New post 29 Aug 2003, 06:30
I understand that part.

However, you've also taken a factor of 2 from 5^21. How did you get that?

i.e. 2(5^21 * 2^21) = 2*10^n
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 [#permalink] New post 29 Aug 2003, 06:33
no i didn't

the number inside the parenthesis are MULTIPLIED, not added. If they were added then I would need to take a factor of two from them both, but since they are multiplied i do not have to do so.

(4*8) = 2(4*4)
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 [#permalink] New post 29 Aug 2003, 06:35
Okay i think i see where i am oing astray. I'm thinking of addition rather than mutiplication.

ab + ac = a(b+c) where as ab *cd = a(b*cd)

right?

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 [#permalink] New post 29 Aug 2003, 06:37
perfecto!

now when you see that on test day you'll think of this conversation and NAIL IT!!! :-D
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  [#permalink] 29 Aug 2003, 06:37
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