## Task B-3.1.

Ordered pairs $(a, b)$ and $(c, d)$ are solutions to the system of equations

$$
\begin{aligned}
\log _{225} x+\log _{64} y & =4 \\
\log _{x} 225-\log _{y} 64 & =1
\end{aligned}
$$

Calculate $\log _{30}(a b c d)$.
## Solution.

The solutions $x$ and $y$ of the problem must be positive real numbers different from 1. Notice that

$$
\log _{x} 225=\frac{1}{\log _{225} x} \quad \text{and} \quad \log _{y} 64=\frac{1}{\log _{64} y}
$$

If we introduce the substitutions $\log _{225} x=m$ and $\log _{64} y=n$, the system transforms into

$$
\begin{aligned}
m+n & =4 \\
\frac{1}{m}-\frac{1}{n} & =1
\end{aligned}
$$

From the first equation, express $m$ and substitute it into the second equation. We get

$$
\begin{aligned}
\frac{1}{4-n}-\frac{1}{n} & =1 \\
n-(4-n) & =4 n-n^{2} \\
n^{2}-2 n-4 & =0
\end{aligned}
$$

The solutions to the last equation and the corresponding $m_{1}, m_{2}$ are:

$$
n_{1}=1-\sqrt{5}, \quad n_{2}=1+\sqrt{5}, \quad \text { respectively } \quad m_{1}=3+\sqrt{5}, \quad m_{2}=3-\sqrt{5}
$$

Now we calculate $x=225^{m}$ and $y=64^{n}$. We have

$$
a=x_{1}=225^{3+\sqrt{5}}, \quad b=y_{1}=64^{1-\sqrt{5}}, \quad c=x_{2}=225^{3-\sqrt{5}} \quad \text{and} \quad d=y_{2}=64^{1+\sqrt{5}}
$$

Thus, the final solution is

$$
\begin{aligned}
\log _{30}(a b c d) & =\log _{30}\left(225^{3+\sqrt{5}} \cdot 64^{1-\sqrt{5}} \cdot 225^{3-\sqrt{5}} \cdot 64^{1+\sqrt{5}}\right)=\log _{30}\left(225^{6} \cdot 64^{2}\right) \\
& =\log _{30}\left(15^{12} \cdot 2^{12}\right)=\log _{30} 30^{12}=12
\end{aligned}
$$