Solve the equation

$$
\frac{1}{\log _{\frac{1}{2}} x}+\frac{1}{\log _{\frac{2}{3}} x}+\cdots+\frac{1}{\log _{\frac{9}{10}} x}=1
$$

on the set of real numbers. ()

Mathematics Competition of Secondary Schools in Hajdú-Bihar County 2002/2003
Solution. Using the known identities of logarithms, we convert the denominator of each term in the sum (1) to base 10 logarithms:

$$
\begin{aligned}
\log _{\frac{1}{2}} x & =\frac{\lg x}{\lg 1-\lg 2}=\frac{\lg x}{-\lg 2} \\
\log _{\frac{2}{3}} x & =\frac{\lg x}{\lg 2-\lg 3} \\
& \vdots \\
\log _{\frac{9}{10}} x & =\frac{\lg x}{\lg 9-\lg 10}=\frac{\lg x}{\lg 9-1}
\end{aligned}
$$

Substituting the reciprocal values into (1), we get:

$$
\frac{-\lg 2}{\lg x}+\frac{\lg 2-\lg 3}{\lg x}+\ldots+\frac{\lg 9-1}{\lg x}=1
$$

We can perform the simplifications, with the common denominator $\lg x \neq 0$, i.e., $x \neq 1$, so the equation is $\lg x=-1$, from which $x=\frac{1}{10}$ is the solution.