Solving a logarithmic equation \ln(x+1)=2+\ln(x-1); solve for x. From there I get \ln

kadetskihykw

kadetskihykw

Answered question

2022-04-22

Solving a logarithmic equation
ln(x+1)=2+ln(x1); solve for x.
From there I get
lnx+1x1=2.

Answer & Explanation

Bethany Mills

Bethany Mills

Beginner2022-04-23Added 14 answers

You are right. From
lnx+1x1=2.
we can write it as
elnx+1x1=e2
then
x+1x1=e2
You can find the solution for x
Jayla Matthews

Jayla Matthews

Beginner2022-04-24Added 18 answers

Yes, you use the fact that
lnu=vu=ev.
So applying this to what you have (with u=x+1x-1 and v=2 ):
(1)   x+1x-1=e2 (1)
Multiplying both sides by x1 gives
(2)   x+1=e2(x1). (2)
(note, here that x=1 is not a solution of (2); so (1) and (2) are equivalent equations)
Solving for x in (2):
x+1=e2xe2
xe2x=1e2
x(1e2)=1e2
x=-1-e21-e2
Or
x=e2+1e2-1.
When solving logarithmic equations, you sould always check your answers. In particular, check that you don't wind up taking the logarithm of a non-positive quantity.

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