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Solve for n. Exponent question

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by krubinoff » Sun Apr 21, 2013 2:54 pm
If 5^21 x 4^11 = 2 x 10^n what is the value of n?

A) 11
B) 21
C) 22
D) 23
E) 32

Can someone assist with this one? Thank you :)
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by guerrero » Sun Apr 21, 2013 3:18 pm
4^11 can be written as (2^2)^11 = 2^22
10^n can be written as (5x2)^n = 5^n x 2^n

so , 5^21 x 2^22 = 2x 2^n x 5^n
= 2^n+1 x 5^n

now you can safely say that n=21

OA [spoiler]21

Hope its clear[/spoiler]
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by krubinoff » Sun Apr 21, 2013 3:32 pm
Gotcha! Perfect. Thank you :)
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by guerrero » Sun Apr 21, 2013 3:37 pm
krubinoff wrote:Gotcha! Perfect. Thank you :)
if you think the answer makes sense , you can hit the Thanks icon as well ;)
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by Anju@Gurome » Sun Apr 21, 2013 6:27 pm
krubinoff wrote:If 5^21 x 4^11 = 2 x 10^n what is the value of n?
(5^21)*(4^11) = (5^21)*(2^22) = 2*(5^21)*(2^21) = 2*[(2*5)^21] = 2*(10^21)

So, n must be equal to 21.

The correct answer is B.
Anju Agarwal
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by Anju@Gurome » Sun Apr 21, 2013 9:04 pm
The easiest solution for this problem is to identify that only source of 5 in the right-hand side of the equation is 10^n. So, number of 5s in the right-hand side is n.

And, number of 5s in the left-hand side is 21.

So, n = 21

The correct answer is B.
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Backup Methods : General guide on plugging, estimation etc.
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