here it goes

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

[(2*2*3)^x] * [2^2(2x+1)]=(2^k) (3^2)
-->(2^2x) * (3^x)*(2^4x+2)=(2^k) (3^2)
--> equatin for x and k

first powers of 3
we get
3^x=3^2

where x=2


now for powers of 2
we get
2x+4x+2=k
6x+2=k

but we know x=2

6(2)+2=14

k=14

Vishu

I'm assuming, it is 4^(2x+1).

Now, split 12^x into 4^x*3^x. The left hand side of the given equation will become:

4^x*3^x*4^(2x+1) i.e. 4^x*4^(2x+1)*3^x. Simplify this to:

4^(3x+1)*3^x (a^x*a^y = a^(x+y))

Simplify 4^(3x+1) to 2^(6x+2) ( (a^x)^2 is a^(2x)).

Now equate 2^(6x+2)*3^x = 2^k*3^2 => x=2 k=6x+2. k=14. E

Damn it. I missed the whole 3^x = 3^2 part! I thought it was simply a bad problem. Thanks a lot guys!