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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.

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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:enter image description here

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 $$$\log_a(x^2-3x)\gt \log_a(4x-x^2) \iff \frac72 \lt x \lt 4$$

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  • $\begingroup$ if $0 < a < 1$ is decreasing... $\endgroup$
    – sirfoga
    Apr 27, 2014 at 16:40
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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.

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  • $\begingroup$ @Macavity Please see my edit. $\endgroup$
    – user122283
    Apr 27, 2014 at 16:43

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