Let $(x_0, y_0)$ be the solution of the following equations $(2x)^{\ln 2} = (3y)^{\ln 3} ; 3^{\ln x} = 2^{\ln y}$. Then $x_0$ is

Let (x₀, y₀) be the solution of the following equations (2x)^(ln 2) = (3y)^(ln 3) ; 3^(ln x) = 2^(ln y). Then x₀ is $(2x)^{\ln 2} = (3y)^{\ln 3}$ $3^{\ln x} = 2^{\ln y}$ $\text{To find } x_0$ $(1)\ \frac{1}{6} \qquad (2)\ \frac{1}{3} \qquad (3)\ \frac{1}{2} \qquad (4)\ 6$ $1.\ \text{take } \ln \text{ on both equations}$ $2.\ \text{set } p=\ln 2,\ q=\ln 3,\ u=\ln x,\ v=\ln y$ $3.\ \text{eliminate } v,\ \text{solve for } u,\ \text{then } x=e^u$ $(\ln x)(\ln 3) = (\ln y)(\ln 2)$ $p = \ln 2, \quad q = \ln 3, \quad u = \ln x, \quad v = \ln y$ $q u = p v \qquad \cdots (A)$ $(\ln 2)(\ln 2 + \ln x) = (\ln 3)(\ln 3 + \ln y)$ $p(p+u) = q(q+v) \qquad \cdots (B)$ $v = \frac{q}{p}\,u$ $p(p+u) = q\left(q + \frac{q}{p}u\right)$ $p^2 + p u = q^2 + \frac{q^2}{p}u$ $u\left(p - \frac{q^2}{p}\right) = q^2 - p^2$ $\frac{p^2-q^2}{p}\,u = -(p^2-q^2) \implies u = -p$ $u = \ln x = -p = -\ln 2 = \ln\frac{1}{2}$ $x = e^{u} = \frac{1}{2}$ $v = \frac{q}{p}(-p) = -q \implies y = \frac{1}{3}$ $(2x)^{\ln 2} = 1^{\ln 2} = 1 = 1^{\ln 3} = (3y)^{\ln 3}\ \ \checkmark$ $\therefore\ x_0 = \frac{1}{2} \Rightarrow \boxed{\text{option (3)}}$