Exponential and Logarithm Equations — Question 5

Question 5

Solve the equation: $$\ln(x - 1) + \ln(x + 1) = \ln(6)$$

• Solve for $x$ algebraically.

• Identify and exclude any extraneous solutions.

Question 5 - Solution

(a) Combine the logarithmic expressions using properties:

$$\ln(x - 1) + \ln(x + 1) = \ln[(x - 1)(x + 1)] = \ln(x^2 - 1)$$

So the equation becomes: $$\ln(x^2 - 1) = \ln(6) \Rightarrow x^2 - 1 = 6 \Rightarrow x^2 = 7 \Rightarrow x = \pm \sqrt{7}$$

(b) Domain Check:

We must have: $$x - 1 > 0 \Rightarrow x > 1 \quad\text{and}\quad x + 1 > 0 \Rightarrow x > -1 \Rightarrow \text{Combined: } x > 1$$

So $x = \sqrt{7} \approx 2.6458$ is valid, but $x = -\sqrt{7} \approx -2.6458$ is not.

$$\text{Final Answer: } \boxed{x = \sqrt{7}}$$