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please explain [#permalink] New post 17 Mar 2007, 21:29
Hello I encountered this problem and I am not even sure how to start it.

5^21 * 4^11 = 2 * 10^n. What is the value of n?
A) 11 B) 21 C) 22 D) 23 E)32
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 [#permalink] New post 17 Mar 2007, 22:35
Hi,


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

you have to further break the equation

it will be come 5^21 * 2^22 (since 2^2 = 4)

or you may also write it as 2*2^21

Now putting this value you will get 2*5^21*2^21 = 2*10^n

or 2* (5*2)^21 = 2* 10^n

or 10^21 = 10^n or n = 21

regards,

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 [#permalink] New post 17 Mar 2007, 22:38
I think this is right...(5^21)*(4^11)=(5^21)*(2^22)=(5^21)*(2)*(2^21)=2*(10^21)
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 [#permalink] New post 18 Mar 2007, 08:23
Hello,

Thank you all so much.
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Re: please explain [#permalink] New post 18 Mar 2007, 11:05
umuhanga wrote:
Hello I encountered this problem and I am not even sure how to start it.

5^21 * 4^11 = 2 * 10^n. What is the value of n?
A) 11 B) 21 C) 22 D) 23 E)32


The general rule is to break it down to prime factors and use the rules of exponents

LHS = 5^21 x 4^11
= 5^21 x [2^2]^11
= 5^21 x 2^22

RHS = 2 x 10^n
= 2 x [2 x 5]^n
= 2 ^ (n + 1) x 5^n


Comparing u get n = 21
Re: please explain   [#permalink] 18 Mar 2007, 11:05
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