Example 14. Solve the equation:
a) $2 \log _{2} x+\log _{\sqrt{2}} x+\log _{1 / 2} x=9$;

b) $\log _{\sqrt{2}}^{2} x+3 \log _{2} x+\log _{1 / 2} x=2$;

c) $\log _{5}^{2} x+\log _{5 x} \frac{5}{x}=1$
Solution. a) The domain of the equation is the interval $x>0$. Since

$$
\begin{aligned}
& \log _{\sqrt{2}} x=\frac{\log _{2} x}{\log _{2} \sqrt{2}}=2 \log _{2} x \\
& \log _{1 / 2} x=\frac{\log _{2} x}{\log _{2} \frac{1}{2}}=-\log _{2} x
\end{aligned}
$$

the original equation is equivalent to the equation

$$
2 \log _{2} x+2 \log _{2} x-\log _{2} x=9
$$

i.e., the equation

$$
\log _{2} x=3 \Leftrightarrow \log _{2} x=\log _{2} 8 \Leftrightarrow x=8
$$

The number 8 is within the domain of the original equation; therefore, it is its only root.

b) The domain of the equation is the set of all positive numbers. We will convert the logarithms in the equation to base $\sqrt{2}$. Since

$$
\begin{aligned}
& \log _{2} x=\frac{\log _{\sqrt{2}} x}{\log _{\sqrt{2}} 2}=\frac{\log _{\sqrt{2}} x}{2}, \\
& \log _{1 / 2} x=\frac{\log _{\sqrt{2}} x}{\log _{\sqrt{2}} \frac{1}{2}}=\frac{-\log _{\sqrt{2}} x}{2}
\end{aligned}
$$

the original equation is equivalent to the equation

$$
\log _{\sqrt{2}}^{2} x+\frac{3}{2} \log _{\sqrt{2}} x-\frac{\log _{\sqrt{2}} x}{2}=2
$$

Let $t=\log _{\sqrt{2}} x$; then the last equation can be written as

$$
t^{2}+3 t / 2-t / 2=2
$$

It is equivalent to

$$
t^{2}+t-2=0 \Leftrightarrow(t+2)(t-1)=0 \Leftrightarrow\left[\begin{array}{l}
t=1 \\
t=-2
\end{array}\right.
$$

Thus, the given equation on its domain is equivalent to the system of two equations:

$$
\left[\begin{array}{l}
\log _{\sqrt{2}} x = 1 \\
\log _{\sqrt{2}} x = -2
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=\sqrt{2} \\
x=1 / 2
\end{array}\right.\right.
$$

Therefore, the set of all solutions to the original equation consists of the numbers $1 / 2$ and $\sqrt{2}$.

c) The domain of the equation is the set $x>0, x \neq 1 / 5$.

We will convert the logarithms in the equation to base 5. Since

$$
\log _{5 x} \frac{5}{x}=\frac{\log _{5} \frac{5}{x}}{\log _{5} 5 x}=\frac{1-\log _{5} x}{1+\log _{5} x}
$$

the original equation is equivalent to the equation

$$
\log _{5}^{2} x+\frac{1-\log _{5} x}{1+\log _{5} x}=1
$$

Let $t=\log _{5} x$, and write the last equation as follows:

$$
t^{2}+\frac{1-t}{1+t}=1
$$

We have:

$$
\begin{aligned}
\frac{t^{2}(t+1)+1-t-(1+t)}{1+t}=0 \Leftrightarrow & \frac{t^{3}+t^{2}-2 t}{t+1}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{t(t-1)(t+2)}{t+1}=0 \Leftrightarrow\left[\begin{array}{l}
t=0 \\
t=1 \\
t=-2
\end{array}\right.
\end{aligned}
$$

Thus, the original equation on its domain is equivalent to the system of three equations:

$$
\left[\begin{array}{l}
\log _{5} x = 0 \\
\log _{5} x = 1, \\
\log _{5} x = -2
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=1 \\
x=5 \\
x=1 / 25
\end{array}\right.\right.
$$

Therefore, the set of all solutions to the given equation consists of three numbers: $1, 5, 1 / 25$.

Using various logarithmic formulas in the process of transforming the equation can lead to both the appearance of extraneous roots and the loss of roots of the original equation.

For example, the transformation of exponentiation of the equation (i.e., replacing the equation $\log _{a} f(x)=\log _{a} g(x)$ with the equation $f(x)=g(x)$) can lead to an expansion of the domain of the original equation and thus to the appearance of extraneous roots. Thus, the equation $f(x)=g(x)$ is a consequence of the equation $\log _{a} f(x)=\log _{a} g(x)$.

110 CH. 2. EQUATIONS WITH ONE UNKNOWN