How do you solve #ln(x+1) - 1 = ln(x-1)#?

1 Answer
Jul 10, 2015

I found: #x=-(1+e)/(1-e)#

Explanation:

I would rearrange your equation as:
#ln(x+1)-ln(x-1)=1#
now I use the fact that:
#loga-logb=log(a/b)#
to get:
#ln((x+1)/(x-1))=1#
we can take the exponential of both sides:
#e^(ln((x+1)/(x-1)))=e^1# eliminating the #ln#;
#(x+1)/(x-1)=e#
#x+1=ex-e#
rearranging:
#x-ex=-1-e#
#x(1-e)=-(1+e)#
#x=-(1+e)/(1-e)#