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

1 Answer
Ken C.
Jun 26, 2016

Do some algebra and apply some log properties to get x=-(e+1)/(1-e)~~2.164x=−e+11−e≈2.164

Explanation:

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

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

Raise both sides to the power of ee:
e^ln((x+1)/(x-1))=e^1eln(x+1x−1)=e1
->(x+1)/(x-1)=e→x+1x−1=e

This equation is solved in a few steps:
x+1=e(x-1)x+1=e(x−1)
x+1=ex-ex+1=ex−e
x-ex=-e-1x−ex=−e−1
x(1-e)=-e-1x(1−e)=−e−1
x=-(e+1)/(1-e)~~2.164x=−e+11−e≈2.164

Note that (-e-1)/(1-e)=(-1(e+1))/(1-e)=-(e+1)/(1-e)−e−11−e=−1(e+1)1−e=−e+11−e.

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