rewrite it as

2^2x * 3^x * 4^2x+1 = 2^k * 3^2
2^6x+2* 3^x = 2^k * 3^2
so x = 2

6x+2 = k
therefore k =14

If x and k are integers and (12^x)(4^2x+1) = (2^k) (3^2), what is the value of K. ( i am assuming 4^(2x+1) i.e 4 to the power of (2x+1)

12^x=(2*2*3)^x=2^2x *3^x
4^2x+1=(2^2)^2X*2^2=2^(4x+2)
so we have now
(12^x)(4^2x+1) = (2^k) (3^2)
(2^2x *3^x)(2^(4x+2))=(2^k) (3^2),
2^(2x+4x+2).3^x=(2^k) (3^2)
L.H.S=R.H.S
so we can say 2^(2x+4x+2).= 2^k ----equation 1
3^x= (3^2) --- equation 2
since both base is 3 we can conclude x=2
now replacing value of x in equation 1
2^(2x+4x+2).= 2^k
since base is 2
we can equate 2x+4x+2=k
6x+2=k
6*2+2=k now replacing value of x
hence k=14