Given matrices A = [[k, 4], [3, 2]] and B = [[1, 2], [0, 4]], find the value of k such that the determinant of the product AB is 4.

Given matrices:

$$A = \begin{pmatrix} k & 4 \\ 3 & 2 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 1 & 2 \\ 0 & 4 \end{pmatrix}$$

Multiply $A$ by $B$:

$$AB = \begin{pmatrix} k & 4 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 4 \end{pmatrix}$$ $$AB = \begin{pmatrix} (k)(1) + (4)(0) & (k)(2) + (4)(4) \\ (3)(1) + (2)(0) & (3)(2) + (2)(4) \end{pmatrix}$$ $$AB = \begin{pmatrix} k + 0 & 2k + 16 \\ 3 + 0 & 6 + 8 \end{pmatrix}$$ $$AB = \begin{pmatrix} k & 2k + 16 \\ 3 & 14 \end{pmatrix}$$

For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is $ad - bc$.

$$\det(AB) = (k)(14) - (2k + 16)(3)$$ $$\det(AB) = 14k - (6k + 48)$$ $$\det(AB) = 14k - 6k - 48$$ $$\det(AB) = 8k - 48$$

We are given that $\det(AB) = 4$.

$$8k - 48 = 4$$

Add 48 to both sides of the equation:

$$8k = 4 + 48$$ $$8k = 52$$

Divide by 8:

$$k = \frac{52}{8}$$

Simplify the fraction:

$$k = \frac{13}{2}$$ $$k = 6.5$$

The value of $k$ is $\boxed{6.5}$.