Logarithmic Equations with Extraneous Solutions Practice Test

Advanced Algebra Practice Test: ACT math skills.

Logarithmic Equations with Extraneous Solutions Practice Test

This test has 20 questions

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Extraneous Solution Screening

Solve the algebra. Inspect every candidate.

A logarithmic equation is not finished when algebra produces roots. Every candidate must pass the original positivity conditions and must make the original equation true.

Domain firstCandidate listOriginal check

Write restrictionsRequire each original argument to be positive.

Solve carefullyKeep every algebraic candidate.

Screen candidatesReject values outside the original domain.

Substitute backConfirm equality in the starting equation.

What makes a solution extraneous?

An extraneous value solves a transformed equation but not the original logarithmic equation.

Definition

Algebraic candidate

A value obtained after combining logarithms, converting to exponential form, factoring, or taking roots.

Actual solution

A candidate that keeps every original logarithm defined and satisfies the original equality.

Key distinction: algebra generates candidates; the original equation decides which candidates are solutions.

The original domain is the first checkpoint

Every logarithm argument must be positive before any product or quotient rule is used.

Domain passport

\log_{b} (F (x)) requires F (x) > 0

One logarithm

F (x) > 0

Two logarithms

F (x) > 0 and G (x) > 0

Valid base

b > 0, b \neq 1

Do not replace separate restrictions with product positivity. A positive product can come from two negative factors, but neither negative factor is a valid logarithm input.

Worked example: one accepted root and one extraneous root

Write the shared domain before condensing the two logarithms.

Full inspection

\log_{2} (x - 1) + \log_{2} (x - 3) = 3

Domain

x > 3

Both original arguments must be positive.

Condense

\log_{2} ((x - 1) (x - 3)) = 3

Addition of logs becomes multiplication inside.

Convert

(x - 1) (x - 3) = 8

Exponential form removes the logarithm.

Factor

(x - 5) (x + 1) = 0

The candidates are five and negative one.

✓

Accepted

$x = 5$ satisfies the domain and the original equation.

Rejected

$x = - 1$ makes both original arguments negative.

Worked example: an even-power equation creates two candidates

A square root step produces both signs, but the domain keeps only one.

Root screening

\log_{3} (x + 2) + \log_{3} (x - 2) = 2

Domain

x > 2

The stricter original inequality controls both logs.

Condense

(x + 2) (x - 2) = 9

Convert the combined logarithm to exponential form.

Solve

x^{2} = 13, x = \pm \sqrt{13}

Both algebraic signs are candidates.

✓

Accepted

$x = \sqrt{13}$ is greater than two.

Rejected

$x = - \sqrt{13}$ fails the original domain.

A domain contradiction can prove there is no real solution

Check whether all original positivity conditions can hold at the same time.

Entry denied

\log_{4} (x - 2) = \log_{4} (1 - x)

First input

x > 2

Second input

x < 1

Intersection

\emptyset

No real value satisfies both restrictions. Setting the arguments equal would produce

x = \frac{3}{2}

, but that value is outside both required intervals.

Keep the original intersection after condensation

A transformed product can be positive on intervals that the separate logarithms never allowed.

Boundary map

Separate inputs control the domain

\log (x - 5) + \log (x + 1)

x > 5

The product is also positive below negative one, but those values make both original logarithm inputs negative.

Equal logarithms still require a domain check

Matching arguments is valid only after both original logarithms are known to exist.

Same-base file

\log_{5} (x + 4) = \log_{5} (2 x - 1)

Domain

x > \frac{1}{2}

The second argument gives the stricter condition.

Match inputs

x + 4 = 2 x - 1

Same-base logarithms form a one-to-one function.

Solve

x = 5

The candidate satisfies both original inputs.

A quotient equation needs denominator and log-domain checks

Preserve the restrictions from both original logarithms before solving the rational equation.

Two-part check

\log_{2} (x + 3) - \log_{2} (x - 1) = 1

Domain

x > 1

Both inputs are positive, and the denominator cannot vanish.

Condense

\frac{x + 3}{x - 1} = 2

Convert the single logarithm to exponential form.

Solve

x + 3 = 2 x - 2, x = 5

The candidate is inside the original domain.

The four-checkpoint workflow

Do not wait until the end to think about restrictions.

Screening route

Checkpoint sequence

Write every original positivity inequality.

Transform and solve without discarding algebraic candidates.

Compare each candidate with the original domain.

Substitute surviving values into the original equation.

Skills Covered and How to Approach

Coordinate logarithm properties, algebraic solving, domain intersections, and final verification.

Training desk

Skills Covered

Core abilities developed by this practice test.

Write logarithm domainsSolve every original positivity condition.

Find shared intervalsIntersect restrictions from multiple logs.

Generate candidatesCondense, convert, factor, and take roots.

Detect extraneous valuesCompare roots with the original domain.

Verify final solutionsSubstitute into the starting equation.

How to Approach

A five-step customs procedure.

List original argumentsWrite one positivity inequality for each.

Intersect the conditionsState the allowed interval before transforming.

Solve the transformed equationKeep every algebraic candidate for screening.

Reject domain failuresOne invalid original input is enough to reject.

Substitute and reportGive only values that satisfy the original equation.

Common Mistakes

Most lost points come from treating candidates as final answers.

Violation list

Writing the domain after solving

Restrictions are easiest to preserve when written first.

Checking only the condensed form

Return to every separate original argument.

Keeping product-positive intervals

Each logarithm input must be positive separately.

Discarding a root too early

Keep algebraic candidates until the domain check.

Missing a square-root branch

An even-power equation produces both signs.

Allowing a zero argument

Logarithm inputs must be strictly positive.

Matching arguments with different bases

The one-to-one shortcut requires the same base.

Reporting an intermediate candidate set

Final answers include only verified solutions.

Final solution audit

Inspect the original domain, every algebraic branch, and the starting equality.

Cleared

Candidate screening complete

A correct answer contains every valid solution and no value that violates an original logarithm input.

Domain · Solve · Screen · Verify

Was every original argument required to be positive?Write separate inequalities before combining logs.

Was the full domain intersection recorded?A value must satisfy all original restrictions.

Were all algebraic candidates kept initially?Include both branches from even powers.

Were invalid candidates explicitly rejected?Identify which original condition fails.

Were surviving values substituted into the original?Confirm both definition and equality.

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