Find Solution Variable x0 for System of Logarithmic Equations

To find the value of $x_0$ for the given system of logarithmic equations:

$(2x)^{\ln 2} = (3y)^{\ln 3} \quad \text{--- (1)}$
$3^{\ln x} = 2^{\ln y} \quad \text{--- (2)}$

Taking the natural logarithm ($\ln$) on both sides of equation (2):
$\ln\left(3^{\ln x}\right) = \ln\left(2^{\ln y}\right)$
$\ln x \cdot \ln 3 = \ln y \cdot \ln 2$
$\Rightarrow \ln y = \frac{\ln 3}{\ln 2} \ln x \quad \text{--- (3)}$

Now, taking the natural logarithm ($\ln$) on both sides of equation (1):
$\ln\left((2x)^{\ln 2}\right) = \ln\left((3y)^{\ln 3}\right)$
$\ln 2 \cdot \ln(2x) = \ln 3 \cdot \ln(3y)$
$\ln 2 \cdot (\ln 2 + \ln x) = \ln 3 \cdot (\ln 3 + \ln y)$
$(\ln 2)^2 + \ln 2 \ln x = (\ln 3)^2 + \ln 3 \ln y \quad \text{--- (4)}$

Substitute equation (3) into equation (4):
$(\ln 2)^2 + \ln 2 \ln x = (\ln 3)^2 + \ln 3 \left( \frac{\ln 3}{\ln 2} \ln x \right)$
$(\ln 2)^2 + \ln 2 \ln x = (\ln 3)^2 + \frac{(\ln 3)^2}{\ln 2} \ln x$

Multiplying the entire equation by $\ln 2$:
$(\ln 2)^3 + (\ln 2)^2 \ln x = \ln 2 (\ln 3)^2 + (\ln 3)^2 \ln x$

Rearranging the terms to isolate $\ln x$:
$(\ln 2)^3 - \ln 2 (\ln 3)^2 = (\ln 3)^2 \ln x - (\ln 2)^2 \ln x$
$\ln 2 \left[ (\ln 2)^2 - (\ln 3)^2 \right] = -\ln x \left[ (\ln 2)^2 - (\ln 3)^2 \right]$

Since $\ln 2 \neq \ln 3$, we have $(\ln 2)^2 - (\ln 3)^2 \neq 0$. Dividing both sides by $\left[(\ln 2)^2 - (\ln 3)^2\right]$ gives:
$\ln 2 = -\ln x$
$\ln x = -\ln 2 = \ln\left(\frac{1}{2}\right)$

Therefore:
$x_0 = \frac{1}{2}$

Correct Answer: C ($\frac{1}{2}$)