Express $\log _{x y} z$ as a function of $\log _{x} z$ and $\log _{y} z$.
Preliminary remark. The question can only be examined in the case where the given logarithms are defined, in other words, for the base of the given logarithms, it holds that $x>0, x \neq 1$, and $y>0, y \neq 1$;

for the base of the sought logarithm, in addition to the above, it holds that $x y \neq 1$;

for the sought $z$ number, it also holds that $z>0$; furthermore, $z \neq 1$, since in the case of $z=1$, we know the result is 0 without any data.

Solution I. From the logarithm identities concerning the change of base, it is sufficient to use the "exchange theorem":

$$
\log _{b} a=\frac{\log _{a} a}{\log _{a} b}=\frac{1}{\log _{a} b}
$$

Accordingly, and with other identities:

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

The transformations performed are permissible according to the preliminary remarks.

It is worth mentioning some special cases.

I. If $x=y$, then

$$
\log _{x^{2}} z=\frac{1}{2} \log _{x} z
$$

II. If $x=z$, then

$$
\log _{x y} x=\frac{\log _{y} x}{1+\log _{y} x}
$$

III. If $x=1 / z$, then

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

(noting that $x y \neq 1$ implies $y \neq z$, so the denominator is $\neq 0$).

Zsuzsa Mayer (Pécs, Széchenyi I. Gymn., III. grade) Károly Vladár (Kiskunhalas, Szilády Á. Gymn., III. grade)

Remarks. 1. For those who are not averse to fractions, the most urgent thing is not to "eliminate" the fractions, so for them, the above result is simpler:

$$
\frac{1}{\log _{x y} z}=\frac{1}{\log _{x} z}+\frac{1}{\log _{y} z}
$$

2. Instead of the well-practiced logarithm identities, we can also reach the goal with the definition of the logarithm, and thus it becomes clear that the logarithm identities are merely different forms of the power identities based on the definition.

Solution II. Let's introduce short notations for the given data and the unknown:

$$
\log _{x} z=a, \log _{y} z=b, \log _{x y} z=c
$$

Thus, according to the definition of the logarithm:

$$
z=x^{a}=y^{b}=x y^{c}
$$

from which

$$
x=z^{\frac{1}{a}}, \quad y=z^{\frac{1}{b}}
$$

so

$$
z=\left(z^{\frac{1}{a}} \cdot z^{\frac{1}{b}}\right)^{c}=z^{c\left(\frac{1}{a}+\frac{1}{b}\right)} \quad(\neq 1)
$$

from which, by equating the exponents,

$$
1=c\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{c(a+b)}{a b}, \quad c=\frac{a b}{a+b}, \quad \frac{1}{c}=\frac{1}{a}+\frac{1}{b}
$$