Logarithmic Equations Practice Test

Advanced Algebra Practice Test: ACT math skills.

Logarithmic Equations Practice Test

This test has 20 questions

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Logarithmic Equations · Domain Passage

Transform the equation, then guard the domain.

Solving a logarithmic equation requires two linked ideas: logarithms reverse exponentiation, and every logarithm must have an allowed base and a positive argument. Algebra creates candidates; the original logarithmic equation decides which candidates may pass.

Gate 01

Logarithmic and exponential forms describe the same relationship

Conversion is the main doorway for equations containing one isolated logarithm.

Inverse forms

\log_{b} (y) = x \Leftrightarrow b^{x} = y

Base condition

b > 0

A real logarithm uses a positive base.

Excluded base

b \neq 1

A base of one cannot create an inverse function.

Argument condition

y > 0

The argument must be strictly positive.

Do not skip the domain: zero and negative arguments are not permitted in real logarithmic equations.

Gate 02

Domain restrictions are written before solving

Every logarithmic argument creates its own strict inequality.

Entry clearance

Example restriction

\log_{3} (x - 1)

x - 1 > 0

x > 1

The boundary value itself is excluded.

Gate 03

One isolated logarithm converts directly

Rewrite in exponential form, solve, and verify the argument.

Direct passage

Basic equation

\log_{2} x = 5

x = 2^{5} = 32

The solution has a positive logarithm argument.

Shifted argument

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

x - 1 = 3^{2} = 9

x = 10

Gate 04

Equal logarithms with the same valid base have equal arguments

This shortcut works only after both arguments are restricted to positive values.

Matching passage

\log_{5} (2 x - 3) = \log_{5} 7

Domain

2 x - 3 > 0

The variable argument must be positive.

Equal arguments

2 x - 3 = 7

The logarithm function is one-to-one.

Solution

x = 5

The argument becomes seven, so the candidate is valid.

Gate 05

A sum of logarithms condenses to a product

Condense first, solve the resulting algebraic equation, and filter every candidate.

Product passage

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

Condensed form

\log_{2} [x (x - 2)] = 3

Both original arguments must still remain positive.

Algebra

x^{2} - 2 x = 8

Move to exponential form.

Candidates

x = 4 or x = - 2

The negative candidate violates the domain.

Valid solution

x = 4

Both original arguments are positive.

Gate 06

A difference of logarithms condenses to a quotient

Restrictions come from the original separate arguments, not only from the final fraction.

Quotient passage

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

Condense and convert

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

The original domain requires $x > 2$ .

Solve and verify

x + 1 = 3 x - 6

x = \frac{7}{2}

The candidate is greater than two and passes.

Gate 07

A power inside the argument is not always removable without care

The domain of the original logarithm controls whether positive and negative roots survive.

Power distinction

Power inside the argument

\log_{3} (x^{2}) = 4

x^{2} = 81

x = - 9 or x = 9

Not the same

Multiplier outside the logarithm

2 \log_{3} x = 4

\log_{3} x = 2

x = 9

Why the solution sets differ: the first argument is positive for either nonzero root. The second equation contains

\log_{3} x

, which requires

x > 0

Gate 08

Multiple logarithms can create extraneous algebraic roots

Factoring may produce several candidates, but the original domain remains the final filter.

Root screening

\log x + \log (x - 9) = 1

Domain

x > 9

Both common-log arguments must be positive.

Condense

x (x - 9) = 10

A common logarithm equal to one has argument ten.

Factor

(x - 10) (x + 1) = 0

The candidates are ten and negative one.

Solution

x = 10

Only ten lies in the original domain.

Gate 09

Change of base connects unfamiliar bases to known logarithms

It can compare bases, evaluate with a calculator, or rewrite an equation in a common language.

Base bridge

\log_{b} M = \frac{\log M}{\log b} = \frac{\ln M}{\ln b}

Rewrite the known side

\log_{4} 16 = 2

This value may be recognized directly because the base squared is sixteen.

Solve the equation

\log_{2} (x + 1) = \log_{4} 16

x + 1 = 4 \to x = 3

Gate 10

A variable base has its own restrictions

Solving the exponential equation is not enough; the base must remain positive and unequal to one.

Base inspection

Equation and candidates

\log_{x} 16 = 4

x^{4} = 16

x = - 2 or x = 2

Base filter

x > 0 and x \neq 1

x = 2

The negative algebraic root cannot serve as a real logarithm base.

Gate 11

Conflicting domain conditions can prove there is no solution

Check the intersection of restrictions before performing unnecessary algebra.

Closed passage

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

x > 4 and x < 2

Restrictions conflict \to No real solution

Domain-first shortcut: no real number can satisfy both strict inequalities, so the logarithms can never be simultaneously defined.

Gate 12

A fixed route keeps transformations and restrictions connected

Use the original equation as the final authority.

Passage map

Decision guide

One isolated logarithm usually converts directly.

Several same-base logarithms usually condense first.

Equal same-base logarithms allow equal arguments.

Unfamiliar bases may use change of base.

Every route ends with an original-equation check.

Gate 14

Skills Covered

The test measures transformation, property use, algebra, domain reasoning, and verification.

Clearance skills

Convert inverse formsMove accurately between logarithmic and exponential equations.

Write domain restrictionsRequire every argument to be positive before solving.

Condense logarithmsApply product, quotient, and power properties correctly.

Solve resulting algebraHandle linear, quadratic, and exponential forms.

Work with basesUse equal bases, change of base, and base restrictions.

Filter candidatesReject extraneous roots and recognize no-solution cases.

Gate 15

How to Approach the Test

Separate restrictions, logarithm structure, algebra, and verification.

Five-gate routine

Write all restrictions

Require positive arguments and valid bases from the original equation.

Intersect every condition.

Identify the equation structure

Look for one logarithm, equal logarithms, several same-base logs, or mixed bases.

Choose the shortest valid route.

Apply one property at a time

Condense sums to products, differences to quotients, or convert directly.

Preserve parentheses.

Move to an algebraic form

Use exponential form or equal arguments, then solve without rounding early.

Keep every candidate.

Verify in the original equation

Check arguments, bases, and equality for each candidate.

Report only valid solutions.

Gate 16

Common Mistakes

Most lost points come from invalid property use or skipped domain checks.

Blocked routes

Allowing a zero argument

Logarithm arguments must be strictly positive, not merely nonnegative.

Combining unlike bases

Product and quotient properties require the same logarithm base.

Turning a sum into a sum inside

A sum of logarithms represents a product of arguments.

Losing a denominator restriction

The original quotient arguments must each be positive.

Using the power rule backward carelessly

Moving an even power can hide an absolute-value or domain issue.

Keeping every quadratic root

Algebraic candidates must still satisfy every original restriction.

Final gate

Logarithmic-equation audit

Confirm restrictions, properties, conversion, algebra, candidate filtering, and original-equation verification.

Passage complete

Domain Gate Check

A candidate becomes a solution only after every original logarithm is defined and the original equation is true.

Restrict · Transform · Solve · Verify

Did I write every argument restriction?Each original logarithm needs a strictly positive argument.

Are all logarithm bases valid?Check positivity and exclude a base of one.

Did I use only valid logarithm properties?Match bases and preserve products, quotients, and powers.

Did I keep and then filter all candidates?Do not discard early or accept an extraneous root.

Does each result satisfy the original equation?Check that both sides are defined and equal.

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