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

1 Answer
Alan P.
Feb 24, 2016

#x= (e+1)/(e-1)#

Explanation:

Given
#color(white)("XXX")ln(x+1)-1=ln(x-1)#

#rArr#
#color(white)("XXX")ln(x+1)-ln(x-1) = 1 = ln(e)#

#color(white)("XXX")ln((x+1)/(x-1))= ln(e)#

#color(white)("XXX")(x+1)/(x-1)=e#

#color(white)("XXX")x+1=ex-e#

#color(white)("XXX")x-ex=-e-1#

#color(white)("XXX")x(1-e)=-e-1#

#color(white)("XXX")x=(e+1)/(e-1)~~2.164#

Related questions
See all questions in Logarithmic Models
Impact of this question
1168 views around the world
You can reuse this answer
Creative Commons License