Logarithmic inequality difficulty

Question

Good day everyone, I would need some assistance in solving this logarithmic inequality: $$\ln \frac {x - 1}{x + 1} \lt 1$$

The answer I came up with is: $\frac {e + 1}{1 - e} \lt x \lt -1 \lor x \gt 1$. Could you please give me a hand? Thank you very much.

You say that you have already come up with an answer. What exactly do you need a hand with??? — barak manos, Dec 20 '16 at 16:03
The answer I came up with is not correct. So I needed a hand in finding the right answer. — Samuel James, Dec 20 '16 at 16:08

hamam_Abdallah · Answer 1 · 2016-12-20T17:10:24.137

4

The domain is $(-\infty,-1)\cup(1,+\infty)$.

the inequation is equivalent to $$\frac{x-1}{x+1}<e$$

thus the solution is

$$(-\infty,\frac{1+e}{1-e})\cup(1,+\infty)$$

edited Dec 20 '16 at 17:10

answered Dec 20 '16 at 15:37

hamam_Abdallah

"the inequation is equivalent to..." You need $0$ on the left of that inequality. – zhw. Dec 20 '16 at 16:59
The inequality is equivalent to $0<(x-1)/(x+1) < e.$ I see now that you took care of this in your first line. – zhw. Dec 20 '16 at 17:06
I still don't understand. Why did you have x < (e + 1)/(e -1) v x > 1 as solution. Please explain and thanks in advance. – Samuel James Dec 20 '16 at 17:10
Ok. I get it now @SalahFatima. Thanks. – Samuel James Dec 20 '16 at 17:13
@SamuelJames Thanks too. you made me happy. – hamam_Abdallah Dec 20 '16 at 17:14

lab bhattacharjee · Answer 2 · 2016-12-20T15:52:51.827

0

Hint:

First of all, we need $(x+1)(x-1)>0\implies$ either $x>1$ or $x<-1$

$\implies\dfrac{x-1}{x+1}<e$

If $x+1>0, x-1<e(x+1)\iff x>?$

If $x+1<0, x-1>e(x+1)<\iff x<?$

edited Dec 20 '16 at 15:52

answered Dec 20 '16 at 15:37

lab bhattacharjee

this is the right method but it is not working with my problem. See here: [http://math.stackexchange.com/questions/2201633/how-to-solve-this-logarithmic-inequality-with-another-logarithm-as-its-base][1] – Sid Mar 28 '17 at 06:06

2 Answers2