Logarithm Functions — Question 7

Question 7

Solve the following logarithmic equation: $$2 \log_4(x + 1) - \log_4(x - 3) = 1$$

• Solve the equation algebraically.

• State any domain restrictions and eliminate extraneous solutions.

Question 7 - Solution

(a) Combine the logarithms:

Use the power rule:

$$\log_4(x + 1)^2 - \log_4(x - 3) = \log_4\left( \frac{(x + 1)^2}{x - 3} \right)$$

So the equation becomes:

$$\log_4\left( \frac{(x + 1)^2}{x - 3} \right) = 1 \Rightarrow \frac{(x + 1)^2}{x - 3} = 4^1 = 4$$

Multiply both sides by $x - 3$:

$$(x + 1)^2 = 4(x - 3) \Rightarrow x^2 + 2x + 1 = 4x - 12 \Rightarrow x^2 - 2x + 13 = 0$$

Solve with the quadratic formula:

$$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(13)}}{2} = \frac{2 \pm \sqrt{4 - 52}}{2} = \frac{2 \pm \sqrt{-48}}{2} = \frac{2 \pm 4i\sqrt{3}}{2} = 1 \pm 2i\sqrt{3}$$

(b) Domain restrictions:

$x + 1 > 0 \Rightarrow x > -1$ , $x - 3 > 0 \Rightarrow x > 3$

So domain is: $\boxed{x > 3}$

The resulting quadratic has no real roots, so the original real logarithmic equation has no solution.

$$\boxed{\text{No real solution.}}$$