Parametric inequation... Supppose we have $a$ a real positive number that's not equal to $1$. Solve the following inequation:
$$\log_a(x^2-3x)>\log_a(4x-x^2)$$ If it's known that $x=3.75$ is one solution of it.

Hint:
If $a>1$, then:
$$\log_a(x^2-3x)>\log_a(4x-x^2)\\
\implies x^2-3x>4x-x^2\text{ by exponentiation: $a^{\log_a(x^2-3x)}>a^{\log_a(4x-x^2)}$}\\
\implies 2x^2-7x>0$$
Else, $\log_a(x)$ is decreasing for all $x$. So, the inequality has reversed arguments.
