So the inequality:

$$2 \cdot \log_{\sqrt3}{(1-x)} - \log_\sqrt3 {(3-x)} \lt 2$$

Can be written as:

$$\log_{\sqrt3}{(1-x)^2} - \log_\sqrt3 {(3-x)} \lt 2$$ ?????????????

I have tried both on Wolfram|Alpha and both gave different intervals of x, although the logarithm property of exponent is: $\log_a{b}^n = n \cdot \log_a{b}$

The interval of solutions for the first inequality is: $$\frac{-1-\sqrt{33}}{2} \lt x \lt 1$$

While for the second one: $$\frac{-1-\sqrt{33}}{2} \lt x \lt 1 \quad \lor \quad 1 \lt x \lt \frac{\sqrt{33}-1}{2}$$

I guess the reason is the different field of existence "I am not sure what's it called in English, it's basically $\log_a{f(x)} \qquad f(x) \gt 0$ to make the logarithm valid". Or in other words...

First inequality: $$\begin{cases} \frac{-1-\sqrt{33}}{2} \lt x \lt \frac{\sqrt{33}-1}{2} \lor x \gt 3 & \mbox{Solution of inequality between the arguments of the logarithms}\\ x \lt 1 & \mbox{Condition of the first argument} \\ x \lt 3 & \mbox{Condition of the second argument}\end{cases} $$

Second one: $$\begin{cases} \frac{-1-\sqrt{33}}{2} \lt x \lt \frac{\sqrt{33}-1}{2} \lor x \gt 3 & \mbox{Solution of inequality between the arguments of the logarithms} \\ (1-x)^2 \gt 0 & \mbox{Condition of the first argument} \\ x \lt 3 & \mbox{Condition of the second argument} \end{cases}$$

So it will be: $$ \begin{cases} \frac{-1-\sqrt{33}}{2} \lt x \lt \frac{\sqrt{33}-1}{2} \lor x \gt 3 \\ x \neq 1 \\ x \lt 3 \end{cases} $$

So the question is why that difference happens though I only used a logarithmic property, and how to decide the values of {$x$} to be sure that the logarithm is valid, I mean why can't I choose $(1-x)^2 \gt 0$ in the first one too since they're basically the same?

  • 2
    $\begingroup$ When considering the two solutions (intervals), we use the domain of the original inequality. So $x > 1$ is out. $\endgroup$ – The Chaz 2.0 Oct 16 '15 at 18:54
  • $\begingroup$ @TheChaz2.0 That should be written as an answer. $\endgroup$ – wythagoras Oct 16 '15 at 18:55
  • $\begingroup$ Why is the first one "the one written in the book" is the original? $\endgroup$ – Shady Alfred Oct 16 '15 at 18:56
  • $\begingroup$ That's what original means! The ORIGIN of the inequality is a book. We use the inequality in the form given to us. $\endgroup$ – The Chaz 2.0 Oct 16 '15 at 18:57
  • $\begingroup$ Why? lol and what about properties of logarithms? $\endgroup$ – Shady Alfred Oct 16 '15 at 19:01

Note that for $\log_{\sqrt3}{(1-x)}$ to exist requires that $1-x > 0$, but for $\log_{\sqrt3}{(1-x)^2}$ to exist only requires that $x \ne 1$.

  • $\begingroup$ I noticed that. My question is which one should I use since $2 \cdot \log_\sqrt3 {(1-x)} = \log_\sqrt3{(1-x)^2}$ and why? $\endgroup$ – Shady Alfred Oct 16 '15 at 18:58
  • $\begingroup$ The problem uses the 2, not the $^2$. So that is the one you should use. $\endgroup$ – marty cohen Oct 17 '15 at 21:31

Note that throughout, $x\neq 1$ and $x\neq 3$ applies. $$\begin{align} 2\log_\sqrt{3}(1-x)-\log_\sqrt{3}(3-x)&<2\\ \log_\sqrt{3}\frac{(1-x)^2}{3-x}&<\log_\sqrt{3}3\\ \frac {(1-x)^2}{3-x}&<3\\ \frac {(1-x)^2}{3-x}-3&<0\\ \frac {x^2+x-8}{3-x}&<0\\ \frac {x^2+x-8}{x-3}&>0\\ \frac {(x-\alpha)(x-\beta)}{x-3}&>0\\ \text{where $\alpha=\frac {-1-\sqrt{33}}2=-3.372, \beta=\frac{-1+\sqrt{33}}2=2.372$}&\\ \begin{cases}-3.372<x<2.372\;\; (x\neq 1) \\x>3\end{cases}\quad\blacksquare\end{align}$$


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