Take: \(k^{2y}=jk\)
Divide both sides of the equation by \(k\) to get: \(\frac{k^{2y}}{k}=j\)
Simplify the left side: \(k^{2y-1}=j\)


Now take: \(j^{2x}=k^{2y}\)
Replace \(j\) with \(k^{2y-1}\) to get: \((k^{2y-1})^{2x}=k^{2y}\)
Apply the power of a power law to the left side: \(k^{(2x)(2y-1)}=k^{2y}\)
Since the bases are equal, we can write: \((2x)(2y-1) = 2y\)
Divide both sides of the equation by \(2y-1\) to get: \(2x = \frac{2y}{2y-1}\)
Divide both sides of the equation by \(2\) to get: \(x = \frac{y}{2y-1}\)

Answer: C