Okay so there is this complex problem that I did solve but it took me a long time to solve it. I'm interested in a different approach to make problems like this simpler and fast to solve.


My Try:

Case 1: When $$0<x^2+2x-3<1.$$ Inside this case, 4 more cases arise, $$ \text{a) } x+4>0 \text{ and } x>0$$ $$ \text{b) } x+4>0 \text{ and } x<0$$ $$ \text{c) } x+4<0 \text{ and }x>0$$ $$ \text{d) } x+4<0 \text{ and } x<0$$

Case 2: When $$x^2+2x-3>1.$$ Again the same four sub-cases for modulus arise.

On solving the two cases separately and taking the union, I get the answer.

How can this question and similar others be solved without making too many cases? Please let me know your approach

And thanks for giving this your time.

  • $\begingroup$ You have to consider cases only when $x^2+2x-3>0$. If this is negative or 0 then log does not exist. $\endgroup$ – Maria Mazur Aug 26 '17 at 18:59
  • $\begingroup$ @it was a typo. Corrected now. $\endgroup$ – Tanuj Aug 26 '17 at 19:08
  • $\begingroup$ I'm pretty sure in the second case you meant "$>1$", not "$>0$". I took the liberty of fixing it for you. $\endgroup$ – zipirovich Aug 26 '17 at 19:21
  • $\begingroup$ Thanks mate! Yes that was a stupid typo. $\endgroup$ – Tanuj Aug 26 '17 at 19:22
  • $\begingroup$ I don't mind problems leading to many cases. But having $x$ as variable in the base of a $\log$ as well as under the $\log$ is truly sick! $\endgroup$ – Christian Blatter Aug 27 '17 at 9:42

To be honest, I can't think of any way to significantly simplify and shorten a solution to this inequality. It is indeed one of those not too difficult but very lengthy and tedious problems with lots of cases. However, one place where you should save some time is the list of your subcases: subcase "$\text{c) } x+4<0 \text{ and }x>0$" is impossible as the two conditions contradict each other, so you shouldn't even set it up. It's easier to see on the number line: the two absolute values switch at $x=-4$ and $x=0$ respectively, splitting the number line into three (not four) intervals, so you have three (not four) subcases.

  • $\begingroup$ Yes that's what I was wondering. Thanks for your effort. $\endgroup$ – Tanuj Aug 26 '17 at 19:35

The domain gives $x^2+2x-3>0$, which is $(-\infty,-3)\cup(1,+\infty)$;

$x^2+2x-3\neq1$, which is $\mathbb R\setminus\{-1-\sqrt5,1+\sqrt5\}$ and

$\frac{|x+4|-|x|}{x-1}>0,$ which is $(-\infty,-2)\cup(1,+\infty)$ by the intervals method.

All these things give the domain: $$(-\infty,-1-\sqrt5)\cup(-1-\sqrt5,-3)\cup(1,-1+\sqrt5)\cup(-1+\sqrt5,+\infty).$$ Now, $\frac{|x+4|-|x|}{x-1}=1$ for $x\in\{5,-5\}$.

Thus, we are ready to use the intervals method:

we'll take on the $x$-axis the following numbers: $$-5,-1-\sqrt5,-3,1,-1+\sqrt5,5$$ and after checking of signs in the intervals we'll get the answer: $$(-1-\sqrt5,-3)\cup(-1+\sqrt5,5).$$ Done!

  • $\begingroup$ if you use $log_{b}c = \frac{logc}{logb} $you only have to make sure $ logc>0$ and $logb>0$ and your condition that $logb\ne 1$ seems a bit strange. Am I wrong? $\endgroup$ – Satish Ramanathan Aug 26 '17 at 19:41
  • $\begingroup$ @satish ramanathan I said $b\neq1$. See please better my solution. $\endgroup$ – Michael Rozenberg Aug 26 '17 at 19:43
  • $\begingroup$ Sorry, you are right. $\endgroup$ – Satish Ramanathan Aug 26 '17 at 19:44
  • $\begingroup$ @satish ramanathan The domain of $\log_bc$ is $b>0$, $c>0$ and $b\neq1$, which I used. $\endgroup$ – Michael Rozenberg Aug 26 '17 at 19:45

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