Example 18. Solve the equation

$$
\log _{x / 2} x^{2}-14 \log _{16 x} x^{3}+40 \log _{4 x} \sqrt{x}=0
$$
Solution.

The first method. Let's find the domain of the equation (16). It is defined by the system of inequalities

$$
x>0 ; \quad x \neq 1 / 16 ; \quad x \neq 1 / 4 ; \quad x \neq 2
$$

The equation (16) on this domain is equivalent to the equation

$$
2 \log _{x / 2} x-42 \log _{16 x} x+20 \log _{4 x} x=0
$$

It is easy to see that $x_{1}=1$ is a root of equation (17), and therefore, of equation (16).

Now let $x$ belong to the domain and $x \neq 1$. For such values of $x$, equation (17) is equivalent to the equation

$$
\frac{1}{\log _{x} \frac{x}{2}}-\frac{21}{\log _{x} 16 x}+\frac{10}{\log _{x} 4 x}=0 \Leftrightarrow
$$

$$
\Leftrightarrow \frac{1}{1-\log _{x} 2}-\frac{21}{1+4 \log _{x} 2}+\frac{10}{1+2 \log _{x} 2}=0
$$

By setting $t=\log _{x} 2$, we arrive at the equation

$$
\frac{1}{1-t}-\frac{21}{1+4 t}+\frac{10}{1+2 t}=0
$$

This equation is equivalent to the system

$$
\left\{\begin{array}{l}
(1+4 t)(1+2 t)-21(1-t)(1+2 t)+10(1-t)(1+4 t)=0 \\
(1-t)(1+4 t)(1+2 t) \neq 0
\end{array}\right.
$$

i.e., the system

$$
\left\{\begin{array}{l}
(t+2)(t-1 / 2)=0 \\
(1-t)(1+4 t)(1+2 t) \neq 0
\end{array}\right.
$$

This system has two solutions: $t_{1}=-2$ and $t_{2}=1 / 2$. Therefore, equation (16) on its domain and for $x \neq 1$ is equivalent to the system of two equations:

$$
\log _{x} 2=-2, \quad \log _{x} 2=1 / 2
$$

From this system, we find the numbers $x_{2}=\sqrt{2} / 2$ and $x_{3}=4$, which are solutions to equation (16).

All transitions were equivalent; therefore, the set of all solutions to equation (16) consists of three numbers:

$$
x_{1}=1, \quad x_{2}=\sqrt{2} / 2, \quad x_{3}=4
$$

If the case $x=1$ had not been considered separately when transitioning from equation (16) to equation (18), the root $x=1$ would have been lost.

Indeed, for $x=1$, $\log _{x / 2} x$ exists and equals 0, but the expression $\frac{1}{\log _{x}(x / 2)}$ is meaningless.

Whenever it is necessary to apply the formula for changing the base, it is best to switch to a base equal to some number. Specific examples suggest which base to use. For example, in equation (16), it would be best to switch to base 2.

The second method. The domain of the equation is defined by the system of inequalities:

$$
x>0, \quad x \neq 1 / 16, \quad x \neq 1 / 4, \quad x \neq 2
$$

In equation (16), we switch to logarithms with base 2. On its domain, it is equivalent to the equation

$$
\frac{2 \log _{2} x}{\log _{2} x-1}-\frac{42 \log _{2} x}{\log _{2} x+4}+\frac{20 \log _{2} x}{\log _{2} x+2}=0
$$

By setting $y=\log _{2} x$, we arrive at the equation

$$
\frac{2 y}{y-1}-\frac{42 y}{y+4}+\frac{20 y}{y+2}=0
$$

which is equivalent to the system

$$
\left[\begin{array}{l}
y=0 \\
\frac{1}{y-1}-\frac{21}{y+4}+\frac{10}{y+2}=0
\end{array}\right.
$$

The second equation of this system is equivalent to the equation

$$
\begin{aligned}
\frac{y+4-21(y-1)}{(y-1)(y+4)}+\frac{10}{y+2} & =0 \Leftrightarrow \frac{-2 y^{2}+3 y+2}{(y-1)(y+4)(y+2)}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{-2(y-2)\left(y+\frac{1}{2}\right)}{(y-1)(y+2)(y+4)}=0 \Leftrightarrow\left[\begin{array}{l}
y=-1 / 2, \\
y=2 .
\end{array}\right.
\end{aligned}
$$

Thus, equation (16) on its domain is equivalent to the system of equations

![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-115.jpg?height=185&width=700&top_left_y=1701&top_left_x=675)

The found numbers $x_{1}=1, x_{2}=\sqrt{2} / 2, x_{3}=4$ belong to the domain of equation (16); therefore, they constitute the set of all its solutions.

The application of the main logarithmic identity

$$
a^{\log a b}=b, \quad b>0, \quad a>0, \quad a \neq 1
$$

can lead to the appearance of extraneous roots if the conditions of its applicability are not monitored.