Math stuff and other things...but mostly math.: AIME problem

The solutions to the system of equations:

$\log_{225}{x} + \log_{64}{y} = 4$

$\log_{x}{225} - \log_{y}{64} = 1$

are $(x_1,y_1)$ and $(x_2,y_2)$ . Find $\log_{30}{(x_1x_2y_1y_2)}$ .

Start with some substitution:

Let $\log_{225}{x} = a$
Let $\log_{64}{y} = b$

Our system of equations now becomes:

$a + b = 4$
$\frac{1}{a} - \frac{1}{b} = 1$

Manipulate the first equation to get $b = 4 - a$ and then substitute into the second:

$\frac{1}{a} - \frac{1}{4-a} = 1$

$\frac{(4-a) - (a)}{a(4-a)} = 1$

$4 - 2a = 4a - a^2$

$a^2 - 6a + 4 = 0$

Quadratic equation:

$a = 3 \pm \sqrt{5}$

We can now solve for b:

$a + b = 4$

$3 \pm \sqrt{5} + b = 4$

$b = 1 \pm \sqrt{5}$

Now we can solve for x and y.

$\log_{225}{x} = 3 \pm \sqrt{5}$

$x = 225^{3 + \sqrt{5}}, 225^{3 - \sqrt{5}}$

y is solved for similarly, I won't bother showing all the steps:

$y = 64^{1 + \sqrt{5}}, 64^{1 - \sqrt{5}}$

Now we have to find what the question asked for, which is:

$\log_{30}{(225^{3 + \sqrt{5}})(225^{3 - \sqrt{5}})(64^{1 + \sqrt{5}})(64^{1 - \sqrt{5}})}$

$= \log_{30}{(225^6)(64^2)}}$

Realize that $30 = 2 \cdot 3 \cdot 5$

$(225^6)(64^2) = (15^{12})(8^4) = (3^{12} \cdot 15^{12})(2^{12})$

From our previous statement, we can see this is equal to $30^{12}$ .

Therefore:

$\log_{30}{30^{12}} = \text{12}$

Solved.