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.
On
The domain is $(-\infty,-1)\cup(1,+\infty)$.
the inequation is equivalent to $$\frac{x-1}{x+1}<e$$
in$(-\infty,-1)$, it becomes $x-1>e(x+1) \iff x<\frac{e+1}{1-e}\approx -2.1.$
in $(1,+\infty)$, it becomes $x-1<e(x+1) \iff x>\frac{e+1}{1-e}$.
thus the solution is
$$(-\infty,\frac{1+e}{1-e})\cup(1,+\infty)$$
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 $x+1<0, x-1>e(x+1)<\iff x<?$