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Gmat Prep (Value of K)

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by raju232007 » Fri Oct 03, 2008 11:48 am
prime factorization is the best method to solve exponent related problems

12^x*4^(2x+1)=2^k*3^2
(2*2*3)^x*(2*2)^(2x+1)=2^k*3^2
2^(6x+2)*3^x=2^k*3^2
Comparing the powers we get
k=6x+2 & x=2
k=6(2)+2=14

The ans is E..i.e 14

Let me know if you still have any doubts..
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by dferm » Fri Oct 03, 2008 12:01 pm
Totally confused with how you jumped from (2*2*3)^x *(2*2)^2x+1 to 2^6x+2*3^x
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by raju232007 » Fri Oct 03, 2008 12:22 pm
(2*2*3)^x*(2*2)^(2x+1)=2^k*3^2
(2^x)*(2^x)*(3^x)*(2^2x)*2*(2^2x)*2=2^k*3^2
2^(6x+2)*3^x=2^k*3^2

By Comparing the powers you will get the solution...i.e 14

still got any doubts?
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by cramya » Fri Oct 03, 2008 4:18 pm
I got 14 also.

Get the bases on both sides to match up on lhs and rhs

End up with

3 ^x * 2 ^ 2x * (2 ^4x * 2 ^2) = (3^2) (2^k)

x = 2

Therefore 2 ^ 2x * (2 ^4x * 2 ^2) simplifies to
2 ^ (6x+2) (a^m * a^n * a ^ k = a ^(m+n+k) where m = 2x,n=4x,k=2)

6x+2 = k
= 6(2)+2 = k

k=14

Choice E)
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