8. Solve the equation $\log _{5}(3 x-4) \cdot \log _{5}(7 x-16) \cdot\left(3-\log _{5}\left(21 x^{2}-76 x+64\right)\right)=1$.
Solution. Rewrite our equation in the form

$$
\log _{5}(3 x-4) \cdot \log _{5}(7 x-16) \cdot\left(3-\log _{5}(3 x-4)-\log _{5}(7 x-16)\right)=1
$$

Such a transformation is possible because the solution to the original equation exists only for $x>\frac{16}{7}$. Let $\log _{5}(3 x-4)=a>0, \log _{5}(7 x-16)=b>0$. Note that both logarithms must be positive. Indeed, the positivity of the first logarithm follows from the domain of existence. Suppose that $\log _{5}(7 x-16)<0 \Leftrightarrow \frac{16}{7}<x<\frac{17}{7}$. In this case, two factors in the left-hand side are positive, and one is negative, which is impossible. Thus, we have

$$
a+b+\frac{1}{a b}=3
$$

Apply the Cauchy inequality to the left-hand side

$$
a+b+\frac{1}{a b} \geq 3 \sqrt[3]{a \cdot b \cdot \frac{1}{a b}}=3
$$

Thus, for the solution of our equation, it is necessary that the Cauchy inequality holds as an equality and $a=b=\frac{1}{a b}$.

## Answer: 3.