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..
Gmat Prep (Value of K)
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raju232007
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raju232007
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(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?
(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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cramya
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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)
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)
