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17 Mar 2007, 21:29
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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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,

Amardeep
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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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18 Mar 2007, 08:23
Hello,

Thank you all so much.
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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
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