## Task $6 / 70$

Determine all positive real numbers $x$ and $y$ that satisfy the equation

$$
\frac{\log _{a} y \cdot \log _{x} y}{\left(\log _{2 x} y\right)^{2}}=\frac{9}{2}
$$
Since the number 1 is unsuitable as the base of a logarithmic system and since $\log 1=0$ for any base, $x=1$ and $y=1$ are initially ruled out as solutions. For further investigation, it is useful to reduce all logarithms to the same base. For this purpose, the equation $\log _{a} c=\frac{1}{\log _{c} a}$ is used. Thus,

$$
\frac{\log _{2} y \cdot \log _{x} y}{\left(\log _{2 x} y\right)^{2}}=\frac{\left(\log _{y} 2 x\right)^{2}}{\log _{y} 2 \cdot \log _{y} x}=\frac{\left(\log _{y} 2+\log _{y} x\right)^{2}}{\log _{y} 2 \cdot \log _{y} x}=\frac{\left(1+\frac{\log _{y} x}{\log _{y} 2}\right)^{2}}{\frac{\log _{y} x}{\log _{y} 2}}=\frac{9}{2}
$$

Substituting $a=\frac{\log _{y} x}{\log _{y} 2}$, we get

$$
\frac{(1+a)^{2}}{a}=\frac{9}{2} \rightarrow a^{2}-\frac{5}{2} a+1=0 \rightarrow a_{1}=2 ; a_{2}=\frac{1}{2}
$$

From $a=\frac{\log _{y} x}{\log _{y} 2}$, it follows that $\log _{y} x=a \cdot \log _{y} 2=\log _{y} 2^{a}$ and thus $x=2^{a}$.

It is evident that the equation is independent of $y$, meaning it is satisfied for any positive real $y$ except $y=1$. With the two values of $a$ determined, we get $x_{1}=4$ and $x_{2}=\sqrt{2}$.