Questions: Solving Logarithmic Equations

Question 1

You solve log₂(x) + log₂(x − 6) = 4 and obtain two candidates: x = 8 and x = −2. What is the correct solution set?

• A. Both x = 8 and x = −2, since the algebra produces two valid numbers
• B. x = 8 only, because x = −2 makes the argument log₂(−2) undefined
• C. x = −2 only, because x = 8 makes the equation unbalanced
• D. No real solution exists — the equation is undefined for positive x

Answer: (B)

After condensing to log₂(x(x−6)) = 4 and converting to x(x−6) = 16, the quadratic yields x = 8 and x = −2. But logarithms require strictly positive arguments. At x = −2, the original expressions log₂(−2) and log₂(−8) are both undefined. x = −2 is an extraneous solution — algebraically produced but not valid. x = 8 gives log₂(8) = 3 and log₂(2) = 1, which sum to 4. Only x = 8 is accepted.

Question 2

To solve log₃(2x + 1) + log₃(x) = 2, what is the correct first step?

• A. Set the arguments equal: 2x + 1 = x, then solve the linear equation
• B. Apply the power rule and rewrite as log₃((2x + 1)·x²) = 2
• C. Condense using the product rule: log₃(x(2x + 1)) = 2, then convert to 3² = x(2x + 1)
• D. Convert each log separately: 3² = 2x + 1 and 3² = x, then solve the system

Answer: (C)

When multiple log terms share the same base, condense them first using log properties before converting to exponential form. The product rule gives log₃(x(2x+1)) = 2, which becomes 9 = 2x² + x, or 2x² + x − 9 = 0. Setting arguments equal (option A) only works when you have log_b(A) = log_b(B) — not log = constant. Converting each log separately (option D) misapplies the definition.

Question 3

If solving a logarithmic equation yields x = 7, and plugging back in gives log₅(7 − 7) = log₅(0), then x = 7 is a valid solution.

• T. True
• F. False

Answer: False

log₅(0) is undefined — logarithms require strictly positive arguments. Even though the algebra produced x = 7, it is an extraneous solution that must be rejected. The argument of any logarithm must be greater than zero. This is why domain checking is mandatory for every candidate solution.

Question 4

An equation log_b(A) = log_b(B) can be solved by setting A = B because the logarithm function is one-to-one.

• T. True
• F. False

Answer: True

This is correct. Because log_b is a one-to-one function (strictly increasing for b > 1), equal outputs force equal inputs: log_b(A) = log_b(B) implies A = B. This is the second main strategy for logarithmic equations. Note that you still must verify the resulting solution makes all original log arguments positive — setting A = B can itself produce an extraneous solution.

Question 5

Why must you always check candidate solutions to a logarithmic equation against the original equation, even when the algebra produces a 'clean' numeric answer?

Model answer: Because logarithms are only defined for strictly positive arguments. When combining log terms using properties (e.g., product or quotient rules) and converting to polynomial form, the algebra can produce values that satisfy the polynomial equation but make one or more original log arguments zero or negative. These are extraneous solutions — artifacts of the algebraic manipulation, not true solutions. Only values that keep every log argument strictly positive are valid.

The domain restriction (argument > 0) is intrinsic to logarithms. Once you convert a log equation to a polynomial via condensing and exponentiating, the polynomial's solution set may include points outside the domain of the original logarithmic expressions. There is no shortcut: every candidate answer must be substituted back into the original equation and all argument expressions checked.