4B. Solve the equation

$$
\log _{6}\left(3 \cdot 4^{-\frac{1}{x}}+2 \cdot 9^{-\frac{1}{x}}\right)+\frac{1}{x}=\log _{6} 5
$$
Solution. It is clear that for $x=0$ the equation is not defined. For $x \neq 0$ we introduce the substitution $-\frac{1}{x}=y$, at which the equation takes the form

$$
\begin{aligned}
& \log _{6}\left(3 \cdot 4^{y}+2 \cdot 9^{y}\right)=y+\log _{6} 5 \\
& \log _{6}\left(3 \cdot 4^{y}+2 \cdot 9^{y}\right)=\log _{6} 6^{y}+\log _{6} 5 \\
& \log _{6}\left(3 \cdot 4^{y}+2 \cdot 9^{y}\right)=\log _{6} 5 \cdot 6^{y}
\end{aligned}
$$

If we antilogarithmize, we get

$$
\begin{aligned}
& 3 \cdot 4^{y}+2 \cdot 9^{y}=5 \cdot 6^{y}, \quad /: 9^{y} \\
& 3 \cdot\left(\frac{2}{3}\right)^{2 y}+2-5 \cdot\left(\frac{2}{3}\right)^{y}=0 .
\end{aligned}
$$

We introduce the substitution $\left(\frac{2}{3}\right)^{y}=t$, at which we get $3 t^{2}-5 t+2=0$. The solutions to the last quadratic equation are $t_{1}=1$ and $t_{2}=\frac{2}{3}$.

- For $t_{1}=1$, we get the equation $\left(\frac{2}{3}\right)^{y}=1$, from which we have $y=0$. The equation $-\frac{1}{x}=0$ has no solutions.
- For $t_{2}=\frac{2}{3}$ we get the equation $\left(\frac{2}{3}\right)^{y}=\frac{2}{3}$ which has the unique solution $y=1$. The solution to the equation $-\frac{1}{x}=1$ is $x=-1$.

Thus, the only solution to the equation is $x=-1$.

## 4th Year