Solve the following system of equations:

$$
\log _{2 x} z=3 ; \log _{5 y} z=6 ;(1) \log _{x y} z=\frac{2}{3}
$$
I. solution. The individual equations are meaningful if $x, y$ and $z$ are positive, $x \neq \frac{1}{2}, y \neq \frac{1}{5}$ and $x y \neq 1$. Under these conditions, (1) is equivalent to the following system of equations:

$$
z=(2 x)^{3} ; z=(5 y)^{6} ; z=(x y)^{\frac{2}{3}}
$$

Substituting the first equation into the second: $(2 x)^{3}=(5 y)^{6}$, and after taking the cube root, we get $x=5^{2} \cdot 2^{-1} \cdot y^{2}$. Substitute this and the second equation of (2) into the third equation:

$$
\begin{gathered}
(5 y)^{6}=\left(5^{2} \cdot 2^{-1} \cdot y^{2} \cdot y\right)^{\frac{2}{3}} \\
5^{6} \cdot y^{6}=5^{\frac{4}{3}} \cdot 2^{-\frac{2}{3}} \cdot y^{2} \\
y=5^{-\frac{7}{6}} \cdot 2^{-\frac{1}{6}}=\frac{1}{5 \sqrt[6]{10}}
\end{gathered}
$$

Based on the used relationships,

$$
\begin{gathered}
z=(5 y)^{6}=\frac{1}{10} \\
x=5^{2} \cdot 2^{-1} \cdot y^{2}=5^{-\frac{1}{3}} \cdot 2^{-\frac{4}{3}}=\frac{1}{2 \sqrt[3]{10}}
\end{gathered}
$$

The obtained values satisfy the system of equations because

$$
(2 x)^{3}=\left(\frac{1}{\sqrt[3]{10}}\right)^{3}=\frac{1}{10}=z, \quad(5 y)^{6}=\left(\frac{1}{\sqrt[6]{10}}\right)^{6}=\frac{1}{10}=z
$$

finally

$$
(x y)^{\frac{2}{3}}=\left(5^{-\frac{1}{3}-\frac{7}{6}} \cdot 2^{-\frac{4}{3}-\frac{1}{6}}\right)^{\frac{2}{3}}=5^{-1} \cdot 2^{-1}=z
$$

Thus, the solution is:

$$
x=\frac{1}{2 \sqrt[3]{10}}, \quad y=\frac{1}{5 \sqrt[6]{10}}, \quad z=\frac{1}{10}
$$

József Pásztor (Eger, Szilágyi Erzsébet Gymnasium, III. grade)

II. solution. Since none of the logarithm values are 0, $z$ cannot be 1. Take the reciprocal of all three equations! According to the identity $\frac{1}{\log _{a} b}=\log _{b} a$, the new system of equations is:

$$
\log _{z} 2 x=\frac{1}{3} ; \log _{z} 5 y=\frac{1}{6} ;(3) \log _{z} x y=\frac{3}{2}
$$

Subtract the third equation from the sum of the first two:

$$
\log _{z} 2 x+\log _{z} 5 y-\log _{z} x y=-1
$$

Based on the properties of the logarithmic function,

$$
\log _{z} \frac{2 x \cdot 5 y}{x y}=\log _{z} 10=-1
$$

This means that $10=z^{-1}$, or $z=\frac{1}{10}$. Then, the values of $x$ and $y$ can be calculated from the first and second equations:

$\log _{1 / 10} 2 x=\frac{1}{3}, \quad$ from which $2 x=\frac{1}{\sqrt[3]{10}} \quad$ and $x=\frac{1}{2 \sqrt[3]{10}}$

$\log _{1 / 10} 5 y=\frac{1}{6}, \quad$ from which $5 y=\frac{1}{\sqrt[6]{10}} \quad$ and $y=\frac{1}{5 \sqrt[6]{10}}$.

The obtained values are substituted similarly to the first solution.

Benedek Valkó (Fazekas M. Főv. Gyak. Gymnasium, IV. grade)