Solve the following equation:

$$
\left(\frac{3}{4}\right)^{\lg x}+\left(\frac{4}{3}\right)^{\lg x}=\frac{25}{12}
$$

(Correction of problem 495.)
Solution: Let $\left(\frac{3}{4}\right)^{\lg x}=y$, then our equation is

$$
y+\frac{1}{y}=\frac{25}{12}
$$

which means

$$
12 y^{2}-25 y+12=0
$$

from which

$$
y_{1}=\frac{4}{3} \quad \text { and } \quad y_{2}=\frac{3}{4} \text {. }
$$

Therefore,

$$
\left(\frac{3}{4}\right)^{\lg x_{1}}=\frac{4}{3}, \quad \lg x_{1}=-1, \quad x_{1}=10^{-1}=\frac{1}{10}
$$

or

$$
\begin{aligned}
\left(\frac{3}{4}\right)^{\lg x_{2}}= & \frac{3}{4}, \quad \lg x_{2}=1 \quad x_{2}=10 . \\
& \text { Endre Hammer (Hajdúnánás, Kőrösi Csoma Sándor g. IV. o. t.) }
\end{aligned}
$$