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.
 A: 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.
A: Since the logarithmic function is a increasing function $\forall a \gt 1$ (and decreasing function $\forall 0\lt a \lt 1$), and continuous $\forall a \in \Bbb{R_+}$, you can use exponentiation method:
$$\log_a(x^2-3x)\gt \log_a(4x-x^2)\\
a^{\log_a(x^2-3x)}>a^{\log_a(4x-x^2)}\\
x^2-3x \gt 4x-x^2\\
2x^2-7x\gt0\\
x(2x-7)\gt0$$
$$x\lt 0 \lor x\gt \frac72$$
However you should take into consideration the existence conditions of the logarithmic function:
\begin{cases}
x^2-3x \gt 0\\
4x-x^2 \gt 0
\end{cases}
\begin{cases}
x\lt 0 \lor x\gt 3\\
0 \lt x \lt 4
\end{cases}
$$3 \lt x \lt 4$$
So the final solution is given by the intersection of
\begin{cases}
x\lt 0 \lor x\gt \frac72\\
3 \lt x \lt 4
\end{cases}
$$\frac72 \lt x \lt 4$$
In fact if you consider the function $f(x)=\ln(x^2-3x)-\ln(4x-x^2)$, and you plot it, you get a similar sketch:
where the dotted black lines are the asymptotes of the 2 logarithmic functions, and the black curve is the sketch $\forall 0\lt a \lt 1$. but since there is a solution for $x=3.75$, we must assume that $a \gt 1\Rightarrow $$$\log_a(x^2-3x)\gt \log_a(4x-x^2) \iff \frac72 \lt x \lt 4$$
