Exponential and Logarithm Equations — Question 3

Question 3

Solve the following equation: $$\log_3(x^2 - 4x) = 1 + \log_3(x - 2)$$

• Solve the equation algebraically.

• Identify any extraneous solutions.

Question 3 - Solution

(a) Combine the logarithms:

Move the constant: $$\log_3(x^2 - 4x) - \log_3(x - 2) = 1$$

Apply the quotient rule: $$\log_3\left( \frac{x^2 - 4x}{x - 2} \right) = 1$$

Factor the numerator: $$x^2 - 4x = x(x - 4) \Rightarrow \frac{x(x - 4)}{x - 2}$$

So: $$\log_3\left( \frac{x(x - 4)}{x - 2} \right) = 1 \Rightarrow \frac{x(x - 4)}{x - 2} = 3^1 = 3$$

Multiply both sides: $$x(x - 4) = 3(x - 2) \Rightarrow x^2 - 4x = 3x - 6 \Rightarrow x^2 - 7x + 6 = 0 \Rightarrow (x - 6)(x - 1) = 0 \Rightarrow x = 6,\ 1$$

(b) Check for extraneous solutions:

Logarithms are only defined for positive arguments.

Check domain: $x^2 - 4x > 0 \Rightarrow x(x - 4) > 0 \Rightarrow x < 0 \text{ or } x > 4$ , $x - 2 > 0 \Rightarrow x > 2$

Only values that satisfy both: $x > 4$

So: $x = 6$ → valid , $x = 1$ → not valid (fails both domain conditions)

Final Answer: $$\boxed{x = 6}$$