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If x and k are integers...

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If x and k are integers...

by iq111 » Sun Sep 20, 2009 10:06 pm
Hello:

I am new to the community, and here to stay. I did a search to this question, but nothing came up. Any help on tackling this problem is appreciated.


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

Please note that (2x+1) is all part of the exponent. The answer is E. (Highlight to see)
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by vijay_venky » Sun Sep 20, 2009 11:30 pm
try to get the equation in to basic prime terms and solve

12^x=(2^2x)*(3^x)
4^(2x+1)=2^(4x+2)

Both combined, 2^(6x+2)*(3^x)=(2^k)*(3^2)
so,x=2 and k=6x+2=14.
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by xcusemeplz2009 » Mon Sep 21, 2009 3:54 am
eqn:- 12^x*4^(2x+1)=2^k*3^2

L.H.S= (2^2x)*(3*x)* (2^4x)*2^2
= 2^(6x+2)*3^x=R.H.S=2^k*3^2
equating exponents
for 2 on both side 6x+2=k and x=2 for 3 on both the side
hence k=6*2+2=14

E
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by iq111 » Mon Sep 21, 2009 7:42 am
vijay, xcusemeplz:

Thanks, I knew there was a reason why 3^2 was in exponential form and not given as 9. It is quite a simple problem once one sees the 'trick'.

Thanks again.
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