Exponential and Logarithm Equations — Question 7

Question 7

Solve for all real solutions: $$\log_2(x + 3) + \log_2(x - 1) = 3$$

Instructions: Solve algebraically and check for extraneous solutions.

Question 7 - Solution

We are given: $$\log_2(x + 3) + \log_2(x - 1) = 3$$

Step 1: Combine the logs using the product rule: $$\log_2[(x + 3)(x - 1)] = 3$$

Step 2: Convert from logarithmic to exponential form: $$(x + 3)(x - 1) = 2^3 = 8 \Rightarrow x^2 + 2x - 3 = 8 \Rightarrow x^2 + 2x - 11 = 0$$

Step 3: Solve the quadratic: $$x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-11)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 44}}{2} = \frac{-2 \pm \sqrt{48}}{2} = \frac{-2 \pm 4\sqrt{3}}{2} = \boxed{-1 \pm 2\sqrt{3}}$$

Step 4: Domain check:

We need: $$x + 3 > 0 \Rightarrow x > -3, \quad x - 1 > 0 \Rightarrow x > 1 \Rightarrow \text{So } x > 1$$

Now test the two roots: $x = -1 + 2\sqrt{3} \approx -1 + 3.464 \approx 2.464$ → valid , $x = -1 - 2\sqrt{3} \approx -1 - 3.464 \approx -4.464$ → invalid

Final Answer: $$\boxed{x = -1 + 2\sqrt{3}}$$