Logarithmic Equations Practice Test

Solve logarithmic equations and reject values that violate argument or base restrictions.

Logarithmic Equations Practice Test

20 logarithmic equations with domain checks, exact answers, and worked explanations.

Instant feedback · Worked explanations

Candidate Verification Pipeline

Solve algebraically. Accept only what survives the original logarithms.

Logarithmic equations have two layers: the algebra that produces candidates and the domain restrictions that decide whether those candidates are legal. Write positivity and base conditions first, combine compatible logarithms, convert a single logarithm to exponential form, solve the resulting algebraic equation, and then return to the original equation for the final decision.

RestrictRecord positive-argument and valid-base conditions.

MergeCondense logs with the same base when helpful.

TranslateConvert one logarithm to exponential form or equate arguments.

ScreenTest every algebraic candidate in the original equation.

1. Domain firewall: restrictions come before algebra

A logarithmic equation is solved only on the set where every original logarithm exists.

Positive arguments only

x - 2 > 0 \Rightarrow x > 2

Write this condition before converting or combining logarithms.

Argument rule

x - 3 > 0

is a typical positivity condition.

Base rule

b > 0, b \neq 1

is required whenever a variable may appear as a logarithm base.

Final screeningAn algebraic root is only a candidate until it satisfies every original restriction.

2. Direct conversion: one logarithm can become one exponential equation

Once the domain is recorded, a simple logarithmic equation can be translated directly.

Logarithmic form

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

x - 2 > 0 \Rightarrow x > 2

The logarithm asks which exponent on the base produces the argument.

convert⇄

Exponential form

x - 2 = 3^{2}

x = 11

The solution satisfies the original positive-argument condition.

3. Several logarithms with the same base can be merged before conversion

The product and quotient rules reduce several logarithms to one, but the original domain restrictions remain.

Merge conveyorWrite the domain first, condense the logarithms, convert to exponential form, then solve the resulting algebra.

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

x > 3

1. Condense

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

Added logarithms become a product inside one logarithm.

2. Convert

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

Use exponential form after only one logarithm remains.

3. Solve algebra

x^{2} - 4 x - 5 = 0

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

4. Produce candidates

x = 5, x = - 1

Do not decide validity yet.

4. Quadratic fork: algebra may produce more roots than the logarithm allows

This is why domain checks are not optional.

Candidate survives

x = 5

\log_{2} (4) + \log_{2} (2) = 2 + 1 = 3

The candidate satisfies the original domain and reproduces the right-hand side.

domain screen⇢

Candidate rejected

x = - 1

x > 3

This candidate violates the original requirement that both logarithm arguments be positive.

5. Common and natural logarithmic equations use the same inverse idea

The notation changes, but the equation-solving structure stays the same.

Natural logarithmInverse of the natural exponential.

Equation

\ln (x) = 4

Solution

x = e^{4}

Common logarithmBase ten is implied.

Equation

\log (x + 1) = 2

Solution

x + 1 = 10^{2} \Rightarrow x = 99

6. Equal logarithms with the same valid base can lead to equal arguments

This shortcut is valid only when both logarithmic expressions are defined and have the same base.

Matching logarithms

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

Because the logarithm function is one-to-one on its domain, equal outputs imply equal valid inputs.

one-to-one⇄

Argument equation

x + 4 = 2 x - 1

After solving, return to the original argument conditions.

7. Nested logarithms are solved from the outside inward

Each inverse step removes one logarithm layer.

Inside-out stackStart with the outermost logarithm because its value is explicitly given.

Original equation

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

Remove the outer logarithm

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

The inner logarithm becomes an ordinary logarithmic equation.

Remove the inner logarithm

x = 3^{4} = 81

Then verify that every nested argument is valid.

8. If the variable is the logarithm base, base validity becomes part of the solution

Solving the resulting power equation is not enough; the logarithm base must remain positive and different from one.

Variable-base equation

\log_{x} (16) = 2

x > 0, x \neq 1

Base restrictions are part of the original equation.

Convert, solve, then screen the base

x^{2} = 16

x = 4, x = - 4

x = 4

The negative algebraic root is rejected because a real logarithm base must be positive.

9. Exact logarithmic solutions can naturally contain radicals

A fractional logarithm value corresponds to a fractional exponent in exponential form.

Fractional outputTranslate directly to a rational exponent.

Equation

\log_{2} (x) = \frac{1}{2}

Exact solution

x = 2^{\frac{1}{2}} = \sqrt{2}

10. Graphical verification: the valid solution is an intersection inside the domain

A graph does not replace the algebra, but it makes the domain restriction and surviving intersection visible.

Equation being checked

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

x > 3

The logarithmic left side exists only to the right of the domain boundary. The valid algebraic solution is the intersection with the horizontal target level inside that region.

Left side versus target valueGreen curve = logarithmic left side; red dashed line = right side.

The marked intersection occurs at the valid candidate. The rejected algebraic candidate lies outside the domain and therefore never appears on the graph of the original logarithmic expression.

11. Error analysis: algebraic candidates are not automatically logarithmic solutions

Most wrong answers come from skipping a structural or domain checkpoint.

Logarithm of a sum split illegally

$\log_{2} (x + 3) \neq \log_{2} (x) + \log_{2} (3)$

Zero or negative argument accepted

Every original logarithm argument must remain strictly positive.

Invalid logarithm base accepted

$b > 0, b \neq 1$ must hold whenever the base is variable.

Second algebraic root omitted before checking

Keep all algebraic roots until the domain screen decides which survive.

Argument confused with final variable value

After exponential conversion, solve the resulting algebraic equation completely.

Domain written after manipulations

Write restrictions first so later equivalent-looking steps do not hide them.

Final candidate audit

Before accepting a logarithmic-equation solution, verify both the algebra and the original domain.

Did I write all positive-argument restrictions first?Do this before combining or converting logarithms.

Did I combine only logarithms with compatible bases?Use product and quotient rules only when their structure is valid.

Did I reduce the equation to one logarithm or equal valid logs?Then convert to exponential form or equate arguments.

Did I solve the resulting algebra completely?Keep all roots, including a second quadratic root, until checking.

If the variable is a base, did I enforce base restrictions?The base must be positive and cannot equal one.

Did every candidate survive substitution into the original equation?Only surviving candidates are actual logarithmic solutions.