Answer: 12

Small Hint:

Set $p = \log_{225} x$ and $q = \log_{64} y,$ using $\log_x 225 = \frac{1}{p}$ and $\log_y 64 = \frac{1}{q}$

Big Hint:

You only need $p_1 + p_2$ and $q_1 + q_2,$ since $x_1 x_2 = 225^{p_1 + p_2};$ get them from $p^2 - 6p + 4 = 0$ by Vieta

Solution:

Let $p = \log_{225} x$ and $q = \log_{64} y,$ so $\log_x 225 = \frac{1}{p}$ and $\log_y 64 = \frac{1}{q}.$ The system becomes $p + q = 4$ and $\frac{1}{p} - \frac{1}{q} = 1.$ Substituting $q = 4 - p$ into the second equation and clearing denominators gives $4 - 2p = p(4 - p),$ that is, $p^2 - 6p + 4 = 0.$

The two solutions of the system correspond to the two roots of this quadratic, so by Vieta’s formulas $p_1 + p_2 = 6,$ and then $q_1 + q_2 = 8 - 6 = 2.$ Hence $$\begin{aligned} x_1 y_1 x_2 y_2 &= 225^{p_1 + p_2} \cdot 64^{q_1 + q_2} \\ &= 225^6 \cdot 64^2 \\ &= 15^{12} \cdot 2^{12} \\ &= 30^{12}, \end{aligned}$$ so $\log_{30}\left(x_1 y_1 x_2 y_2\right) = 12.$