`ln(x+1)-ln(x-1)`

can be written as

`ln((x+1)/(x-1))`

so the equation to be solved is

` `

`ln((x+1)/(x-1))=ln2`` `` `

Considering the exponent of both sides, you will have

`(x+1)/(x-1)=2`

This equation can be solved for values of x different from 1.

It is equivalent to

`x+1=2(x-1)`

That is to say

` x+1=2x-2 `

which gives

` -x=-3 `

So the solution of the initial equation is

` x=3`

First I notice that I have a rule that says that the subtraction of natural logs is the same as the natural log of the division of their arguments. This means I can change the original equation to:

`ln((x+1)/(x-1))=ln(2)`

`(x+1)/(x-1)=2`

`(x-1)*((x+1)/(x-1))=2(x-1)`

`x+1=2(x-1)`

`x+1=2x-2`

`-x -x`

`1=x-2`

`+2 +2`

`3=x`

` `

For both logarithms to make sense the inside must be greater than 0. So x>1 because the x-1 in the second log.

Next, lna-lnb = ln(a/b), so

ln(x+1)-ln(x-1)=ln((x+1)/(x-1))=ln2.

ln is a monotone function, so lna=lnb implies a=b. Therefore

(x+1)/(x-1)=2, or

x+1=2x-2, or x=3. This x is in the domain (>1) so is the correct answer.

First apply the Quotient Rule for logarithms

ln(M)-ln(N) = ln(M/N)

So ln(x+1) - ln(x-1) = ln[(x+1)/(x-1)] = ln(2)

Since the ln(M) = ln(N) tells us that M=N we can write

(x+1)/(x-1) = 2

Multyplying both sides by (x-1) we get

x+1 = 2(x-1)

By Distributive Property

x+1 = 2x - 2

Subtracting x from both sides

1 = x - 2

Adding 2 to both sides

3 = x

so x = 3 in this case

We can double check our answer by plugging back into the equation

ln(3+1)- ln(3-1) = ln(2)

ln(4) - ln(2) = ln(2)

1.38629461 - 0.693147181 = 0.693147181

thus proving = 3 is correct

ln(x+1)-ln(x-1)=ln2

this is of the form ln(a)-ln(b)= ln(a/b)

ln(x+1)-ln(x-1)=ln2

=>ln((x+1)/(x-1))=ln2

=>(x+1)/(x-1)= 2

x+1=2(x-1)

x+1=2x-2

=> x = 3

so when x= 3 then ln(x+1)-ln(x-1)=ln2