The Logarithm That Lost Its Domain
A student solves the equation log₂(x − 1) + log₂(x − 3) = 3 by combining the left side into log₂[(x − 1)(x − 3)] = 3, then writing (x − 1)(x − 3) = 8. Expanding gives x² − 4x + 3 = 8, so x² − 4x − 5 = 0, which factors as (x − 5)(x + 1) = 0. The student reports x = 5 and x = −1 as the full solution set. A classmate substitutes x = −1 into the original equation and gets an error on the calculator. The teacher asks the class to decide who is right and why, and then to solve a related inequality.
1. Determine the correct solution set of the original equation and explain precisely where the student's method went wrong.
Read the worked answer
The original equation requires x − 1 > 0 and x − 3 > 0, i.e. x > 3.
Solving the combined equation gives x = 5 or x = −1. Only x = 5 satisfies x > 3. Check: log₂4 + log₂2 = 2 + 1 = 3. Correct.
x = −1 makes both arguments negative, so neither logarithm is defined over the reals. The combining step log a + log b = log(ab) is valid only when a > 0 and b > 0; applying it outside the domain introduced an extraneous root. The correct solution set is {5}.
2. Solve log₂(x − 1) + log₂(x − 3) ≤ 3 and express the answer as an interval.
Read the worked answer
Domain: x > 3. On this domain the left side equals log₂[(x − 1)(x − 3)], and log₂ is increasing, so the inequality is equivalent to
(x − 1)(x − 3) ≤ 8, i.e. x² − 4x − 5 ≤ 0, i.e. (x − 5)(x + 1) ≤ 0, so −1 ≤ x ≤ 5.
Intersecting with x > 3 gives 3 < x ≤ 5.
Answer: x ∈ (3, 5]. The left endpoint is open because x = 3 makes log₂(x − 3) undefined; x = 5 is included because equality holds there.