What is the exact form and corrected to three decimal places if not an integer of ln (x^2-1)-ln(x-1)=ln6

Solve for x:
log(x^2 - 1) - log(x - 1) = log(6)

log(x^2 - 1) - log(x - 1) = log(1/(x - 1)) + log(x^2 - 1) = log((x^2 - 1)/(x - 1)):
log((x^2 - 1)/(x - 1)) = log(6)

Cancel logarithms by taking exp of both sides:
(x^2 - 1)/(x - 1) = 6

Multiply both sides by x - 1:
x^2 - 1 = 6 (x - 1)

Expand out terms of the right hand side:
x^2 - 1 = 6 x - 6

Subtract 6 x - 6 from both sides:
x^2 - 6 x + 5 = 0

The left hand side factors into a product with two terms:
(x - 5) (x - 1) = 0

Split into two equations:
x - 5 = 0 or x - 1 = 0

Add 5 to both sides:
x = 5 or x - 1 = 0

Add 1 to both sides:
x = 5 or x = 1

log(x^2 - 1) - log(x - 1) ⇒ log(1^2 - 1) - log(1 - 1) = (undefined)
log(6) ⇒ log(6) ≈ 1.79176:
So this solution is incorrect

log(x^2 - 1) - log(x - 1) ⇒ log(5^2 - 1) - log(5 - 1) = log(2) + log(3) ≈ 1.79176
log(6) ⇒ log(6) ≈ 1.79176:
So this solution is correct

The solution is:
Answer: |x = 5

ln (x^2-1) - ln(x-1) = ln(6)      write as

ln [ (x^2 - 1) / (x - 1) ]  =  ln 6     we can forget the logs

(x^2 - 1) / (x - 1)  = 6     rearrange

x^2 - 1  =  6 (x - 1)  simplify

x^2 - 1  = 6x - 6

x^2 - 6x +  5  = 0     factor

(x - 1) ( x -5) = 0    set each factor to 0    and x  = 1  [reject because it makes one or more logs undefined]  or x  = 5  which is valid..

I wonder if we can do it this way?

ln(x^2-1) - ln(x-1) = ln 6

ln ( (x^2-1)/(x-1) ) = ln 6

ln  [ (x-1)(x+1)/(x-1) ] = ln 6

ln  (x+1)  = ln 6

x+1 = 6

x = 5