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

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Intern
Joined: 20 Aug 2003
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5^21 * 4^11= 2* 10^n [#permalink]  25 Aug 2003, 11:39
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$$5^21 * 4^11= 2* 10^n$$

solve for n?
Manager
Joined: 10 Jun 2003
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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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Sept 3rd

Manager
Joined: 14 Aug 2003
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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
Intern
Joined: 17 Aug 2003
Posts: 38
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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

Thanks
Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
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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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Sept 3rd

Intern
Joined: 17 Aug 2003
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Location: USA
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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
Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
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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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Sept 3rd

Intern
Joined: 17 Aug 2003
Posts: 38
Location: USA
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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?

Thanks
Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
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perfecto!

now when you see that on test day you'll think of this conversation and NAIL IT!!!
_________________

Sept 3rd

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