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What is the value of k?!?

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by gmatpup » Tue Nov 08, 2011 7:26 pm
If x and k are integers and (12^x)(4^2x+1) = (2^k)(3^2), what is the value of k?

A. 5

B. 7

C. 10

D. 12

E. 14


I know there is a quick way to solve this, can someone please tell me :)

Answer is: E
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Source: — Problem Solving |

by Anurag@Gurome » Tue Nov 08, 2011 7:41 pm
gmatpup wrote:If x and k are integers and (12^x)(4^2x+1) = (2^k)(3^2), what is the value of k?

A. 5
B. 7
C. 10
D. 12
E. 14

I know there is a quick way to solve this, can someone please tell me :)

Answer is: E
(12^x) * [4^(2x+1)] = (2^k) * (3^2)
(2^2x)*(3^x)* (2^4x) * 2^2 = (2^k) * (3^2)
2^(6x+2) * (3^x) = (2^k) * (3^2)
Since bases are the same, so exponents will be equal.
6x + 2 = k
x = 2
k = (6 * 2) + 2 = 14
The correct answer is E.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
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by neelgandham » Wed Nov 09, 2011 1:34 am
If you ask me, the answer would be 'it is more to do with practice than anything else'. If the same problem is on my math paper(school level). I would have done it this way.

(12^x)*(4^2x+1) = (2^k)*(3^2)
(4^x)*(3^x)*(4^2x+1) = (2^k)*(3^2)
(2^2x)*(3^x)*(4^2x+1) = (2^k)*(3^2)
(2^2x)*(3^x)*(2^2*(2x+1)) = (2^k)*(3^2)
(2^2x)*(3^x)*(2^4x+2) = (2^k)*(3^2)
(2^2x)*(2^4x+2)*(3^x) = (2^k)*(3^2)
(2^(2x+4x+2))*(3^x) = (2^k)*(3^2)
Bases are same, so we can equate the exponents

(3^x) = (3^2), Implies x = 2;
(2^(2x+4x+2))=(2^k), Implies 6x+2 = k; k = (6*2)+2 = 14 (x = 2)

But if the same question is on the GMAT, I would do it the GMAT way:

As you read through the question, your hand(and the pen of course) with the assistance of your brain(again by practice) should give you an equation as shown below by the time you finish reading the question.

2^(6x+2) * (3^x) = (2^k) * (3^2)
x= 2 and 6x+2 = 14 (shouldn't take more than 45 seconds to solve this question in the GMAT exam.
Anil Gandham
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by gmatpup » Wed Nov 09, 2011 6:01 pm
Anurag@Gurome wrote:
gmatpup wrote:If x and k are integers and (12^x)(4^2x+1) = (2^k)(3^2), what is the value of k?

A. 5
B. 7
C. 10
D. 12
E. 14

I know there is a quick way to solve this, can someone please tell me :)

Answer is: E
(12^x) * [4^(2x+1)] = (2^k) * (3^2)
(2^2x)*(3^x)* (2^4x) * 2^2 = (2^k) * (3^2)
2^(6x+2) * (3^x) = (2^k) * (3^2)
Since bases are the same, so exponents will be equal.
6x + 2 = k
x = 2
k = (6 * 2) + 2 = 14
The correct answer is E.



Where did you get x =2 ?

Thanks again :)
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by shankar.ashwin » Wed Nov 09, 2011 8:10 pm
WHen you compare the bases on both sides of the equation, you have 2^(something) and 3^(something)

When 2 equations have same bases, their powers can be equated. SO you have 3^x in the left and 3^2 in the right. So you get x = 2.

Now you can use that to equate powers of 2.
gmatpup wrote:
Where did you get x =2 ?

Thanks again :)
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by gmatpup » Thu Nov 10, 2011 8:10 am
oh wow I totally overlooked that!!!

Thanks sooo much everyone for the clarification, it makes this problem super easy now :)
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