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

1 Answer
Jun 26, 2016

Answer:

Do some algebra and apply some log properties to get #x=-(e+1)/(1-e)~~2.164#

Explanation:

Begin by collecting the #x# terms on one side of the equation and the constant terms (numbers) on the other:
#ln(x+1)-ln(x-1)=1#

Apply the natural log property #lna-lnb=ln(a/b)#:
#ln((x+1)/(x-1))=1#

Raise both sides to the power of #e#:
#e^ln((x+1)/(x-1))=e^1#
#->(x+1)/(x-1)=e#

This equation is solved in a few steps:
#x+1=e(x-1)#
#x+1=ex-e#
#x-ex=-e-1#
#x(1-e)=-e-1#
#x=-(e+1)/(1-e)~~2.164#

Note that #(-e-1)/(1-e)=(-1(e+1))/(1-e)=-(e+1)/(1-e)#.