Strategy: Make one of the expressions with look like another expression.

Take: $j^{2x} = j^4k^2$

Divide both sides of the equation by $j^4$ to get: $\frac{j^{2x}}{j^4} = \frac{j^4k^2}{j^4}$

Simplify: $j^{2x - 4} = k^2$

Raise both sides of the equation to the power of $1.5y$ to get: $(j^{2x - 4})^{1.5y} = (k^2)^{1.5y}$

Simplify: $j^{3xy - 6y} = k^{3y}$

It's given that $j^{2x} = k^{3y}$, we can replace $k^{3y}$ with $j^{2x}$ to get: $j^{3xy - 6y} = j^{2x}$

Since the bases are equal, we can conclude the exponents are equal to get: $3xy - 6y = 2x$

Factor the left side: $y(3x - 6) = 2x$

Solve for y: $y = \frac{2x}{3x - 6}$

Answer: C