log(x + 1) - log(x - 1) < 1/100
-log(x - 1) + log(x + 1) < 1/100
/ 1/100\ -\1 + e / (--------------, oo) 1/100 1 - e
x in Interval.open(-(1 + exp(1/100))/(1 - exp(1/100)), oo)
/ 1/100 \ | 1 + e | And|x < oo, ----------- < x| | 1/100 | \ -1 + e /
(x < oo)∧((1 + exp(1/100))/(-1 + exp(1/100)) < x)