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

Guest Feb 17, 2017

#1**0 **

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**

Guest Feb 17, 2017

#2**0 **

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..

CPhill
Feb 17, 2017