How can I solve the following inequality?

Question

I have the inequality:

$lg((x^3-x-1)^2) < 2 lg(x^3+x-1)$

And I'm not sure how should I go about solving it. I wrote it like this:

$2lg(x^3-x-1) < 2lg(x^3+x-1)$

$lg(x^3-x-1) < lg(x^3 + x - 1)$ (*)

Here I have the conditions:

$x^3-x-1 > 0$

$x^3+x-1 > 0$

At first this stumped me, but then I realised I can just add the inequalities to get:

$2x^3 -2> 0$

$x^3-1>0$

$x^3>1 \Rightarrow x \in (1, + \infty)$

So that is our condition.

Going back to (*) and raising the inequality to the power of $10$, I got:

$x^3-x-1 < x^3+ x - 1$

$2x>0 \Rightarrow x \in(0, +\infty)$

So, if we also consider the condition, we have

$x \in (1, + \infty) \cap (0, + \infty)$

So $x \in (1, + \infty)$.

The problem with this answer is that it is wrong. My textbook lists the following possible answers:

A. $\mathbb{R}$

B. $(0, + \infty)$

C. $(1, + \infty)$

D. $(0, 1)$

E. Other answer

So I got answer C, but after I checked the back of the book I found that the correct answer should be E. So, what did I do wrong and what is that other answer?

Answer 1

The domain it's $$x^3-x-1\neq0$$ and $$x^3+x-1>0.$$ Now, since $$\ln(x^3-x-1)^2=2\ln|x^3-x-1|,$$ we need to solve $$|x^3-x-1|<x^3+x-1$$ or $$-x^3-x+1<x^3-x-1<x^3+x-1.$$ The left inequality gives $x>1$ and the right inequality gives $x>0,$ which gives the answer: $$(1,+\infty)\setminus\{x|x^3-x-1=0\}.$$

Answer 2

Your condition isn't quite right. When $x = 1.1$, $x^3−x−1 = 1.331 - 1.1 - 1 < 0$.

From $\log(x^3 - x - 1) < \log(x^3 + x - 1)$, subtract one side from the other, then use a logarithm property to get one logarithm. \begin{align*} 0 &< \log(x^3 + x - 1) - \log(x^3 - x - 1) \\ 0 &< \log \left( \frac{x^3 + x - 1}{x^3 - x - 1} \right) \end{align*} Apply $f(y) = 10^y$ to both sides to obtain \begin{align*} 1 &< \frac{x^3 + x - 1}{x^3 - x - 1} \text{,} \end{align*} which you demonstrated you knew how to handle in your question. (Don't forget, when you go from $a < \frac{b}{c}$ to $ac < b$ you may have inadvertently multiplied by zero for some values of $x$ that you aren't currently thinking about (yielding $0 < 0$, which is false). You should check what happens in those cases. For instance, here, what happens when $x^3 - x - 1 = 0$ in the original equation?)

Answer 3

Better not get rid of the squares -- they're your friend. So we have $$\log{(x^3-x-1)^2}< \log{(x^3+x-1)^2}.$$ Since the logarithm is monotonic, this implies $$(x^3-x-1)^2< (x^3+x-1)^2,$$ or that $$(x^3-x-1)^2- (x^3+x-1)^2<0.$$ This is easily factored to give $$(x^3-x-1+x^3+x-1)(x^3-x-1-x^3-x+1)<0,$$ or $$(2x^3-2)(-2x)<0,$$ or more simply $$x(x^3-1)=x(x-1)(x^2+x+1)>0,$$ which implies $$x(x-1)>0$$ since $x^2+x+1$ is always positive.

I believe you can now take it from here.

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Well, you also have to factor in that the original inequality makes sense only for $x^3+x-1>0$ and $x^3-x-1\ne 0.$