Question

How do you solve `ln(x+1)-ln(x-2)=ln x`?

Answer 1

`(3+sqrt(13))/2`

Explanation:

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

If:

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

Then:

`(x+1)/(x-2)=x =x^2-3x-1`

Solving: `x^2-3x-1`

`x = (3+sqrt(13))/2 and x= (3-sqrt(13))/2`

`(3-sqrt(13))/2` not valid as gives negative value in `ln(x-2)`

So solution is:

`(3+sqrt(13))/2`

Answer 2

`x = 3/2 +sqrt(13)/2`

Explanation:

Taking the exponent of both sides, we get:

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

Multiplying both sides by `(x-2)` we get:

`x+1 = x^2-2x`

Subtract `x+1` from both sides to get:

`0 = x^2-3x-1`

`color(white)(0) = (x-3/2)^2-9/4-1`

`color(white)(0) = (x-3/2)^2-(sqrt(13)/2)^2`

`color(white)(0) = (x-3/2-sqrt(13)/2)(x-3/2+sqrt(13)/2)`

So:

`x = 3/2 +-sqrt(13)/2`

One of these values `3/2+sqrt(13)/2` is positive and a valid solution of the original equation.

The other value `3/2-sqrt(13)/2` is negative so not a solution if `ln` is the real logarithm. In fact it is not even a solution if we consider the complex logarithm, for which:

`ln(3/2-sqrt(13)/2+1) - ln(3/2-sqrt(13)/2-2)`

`= ln(5/2-sqrt(13)/2) - ln(1/2+sqrt(13)/2)-ipi`

`!= ln(sqrt(13/2)-3/2)+ipi = ln(3/2-sqrt(13)/2)`

Answer 3

See below.

Explanation:

In the quest for real solutions, supposing `x ne {0,2}`

`log_e abs(x+1)-log_e abs(x-2) = log_e absx` or

`log_e(abs(x+1)/(abs(x-2)absx))=0` or

`abs(x+1)/(abs(x-2)absx)=1` whose solutions are included in the solutions for

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

`(x+1-(x-2)x)(x+1+(x-2)x)=0` or

`(1+3x-x^2)(1-x+x^2)=0`

and the real solutions are from

`1+3x-x^2=0` with the only feasible solution

`x = 1/2 (3 + sqrt[13])`