Logarithm Functions — Question 5

Question 5 - Solution

(a) Domain:

We need both logarithmic expressions to be defined: $$x - 1 > 0 \Rightarrow x > 1,\quad x + 3 > 0 \Rightarrow x > -3$$

The more restrictive condition is $x > 1$, so:

$$\boxed{\text{Domain: } (1, \infty)}$$

(b) Use the product rule for logarithms: $$f(x) = \log_5[(x - 1)(x + 3)] = \log_5(x^2 + 2x - 3)$$

(c) Solve the equation: $$\log_5(x^2 + 2x - 3) = 2 \Rightarrow x^2 + 2x - 3 = 5^2 = 25 \Rightarrow x^2 + 2x - 28 = 0$$

Use the quadratic formula: $$x = \frac{-2 \pm \sqrt{(2)^2 + 4 \cdot 28}}{2} = \frac{-2 \pm \sqrt{4 + 112}}{2} = \frac{-2 \pm \sqrt{116}}{2}$$ $$= \frac{-2 \pm 2\sqrt{29}}{2} = -1 \pm \sqrt{29}$$

Check for extraneous solutions:

$$-1 + \sqrt{29} \approx -1 + 5.385 \approx 4.385\ (\text{valid}),\quad -1 - \sqrt{29} < 0\ (\text{invalid})$$

Only the positive root satisfies $x > 1$, so:

$$\boxed{x = -1 + \sqrt{29}}$$