Solve simple logarithmic equations

Ontario Grade 12 Mathematics

Exponential and Logarithmic Functions

Use the meaning of a logarithm, check the domain, and verify the solution

A logarithmic equation contains a logarithm with an unknown value. The key idea is that a logarithm tells you which exponent is needed. For example, asking for

\log_2 8

is the same as asking, “What power of

2

equals

8

?” Since

2^3=8

, the logarithm is

3

. In this lesson, you will use that connection to solve simple equations and make sure each answer is allowed.

What you will learn

Explain a logarithm as an exponent.
Solve simple logarithmic equations by rewriting them in exponential form.
Use the requirement that a logarithm’s argument must be positive.
Check a proposed solution in the original equation.

1. Prerequisite bridge: powers and logarithms

An exponent tells how many times a base is used as a factor. In

2^3=8

, the base is

2

, the exponent is

3

, and the value is

8

. A logarithm reverses this question: it gives the exponent when you know the base and the value.

In

\log_b a=c

, the base is

b

, the argument is

a

, and the logarithm’s value is

c

. This statement means that

b

raised to the power

c

equals

a

. The base must be positive and cannot equal

1

, and the argument must be positive.

These conditions matter when solving equations. A logarithm such as

\log_5(x-2)

is defined only when its argument,

x-2

, is positive. So its input must satisfy

x>2

. This restriction is called a domain condition: it tells which input values are allowed.

\log_b a=c\iff b^c=a

A logarithm answers an exponent question.
The argument of a logarithm must be positive.
The base of a logarithm must be positive and different from $1$ .

2. A reliable method for simple equations

When an equation has one logarithm equal to a number, use the definition to rewrite it as an exponential equation. Then solve the resulting equation using familiar operations. For instance, the structure

\log_b A=c

becomes

b^c=A

, where

A

is the expression inside the logarithm.

Before solving, note the domain condition

A>0

. After finding a possible value for the unknown, check that it makes the argument positive. Then substitute the value into the original logarithmic equation. This check can catch an answer that came from algebra but is not allowed in the original equation.

Some simple equations have a logarithm on each side with the same base. If both logarithms are defined, equal logarithm values with the same base have equal arguments. For example, an equation shaped like

\log_b A=\log_b B

can be changed to

A=B

, while keeping both conditions

A>0

and

B>0

. Solve the resulting equation and test the domain conditions.

\log_b A=\log_b B\iff A=B\quad(A>0,\ B>0)

Rewrite a logarithmic equation as an exponential equation when a logarithm equals a number.
For matching bases on both sides, equate the arguments.
A value is a solution only if it satisfies the original equation and its domain conditions.

3. See the connection in words, numbers, and symbols

Consider the question, “To what power must

3

be raised to make

27

?” The answer is

3

, because

3^3=27

. The same fact can be written as a logarithm statement. These are not separate facts; they are two forms of the same relationship.

A table can help you move between the forms. In the first row, the exponent is unknown in the logarithmic question. In the second row, the exponential statement makes that exponent visible. When solving, you can use the same conversion with an expression in place of the number.

\log_3 27=3\iff 3^3=27

The logarithmic form asks for an exponent.
The exponential form displays the power relationship directly.
Replace the argument with an expression when solving for an unknown.

4. Apply the method and check the result

A solution should pass two checks. First, its value must make each logarithm’s argument positive. Second, substitution into the original equation must make the two sides equal. Do not rely only on a value produced by rearranging equations.

If a logarithm’s argument is a simple expression, its positivity condition can often be found before solving. For an argument such as

2x+1

, require

2x+1>0

. Keep that restriction in mind while solving and checking.

For the final check, evaluate the logarithm using the original base and argument. If the equation says a logarithm equals an integer, converting back to a power is often the clearest way to verify it. State the solution only after it passes both checks.

\text{argument}>0

Record argument restrictions before accepting a value.
Verify by substituting into the original equation.
Write the final solution clearly.

One relationship in two forms

Logarithmic question	Exponential statement	Meaning
$\log_3 27=3$	$3^3=27$	The exponent on $3$ is $3$ .
$\log_2 8=3$	$2^3=8$	The exponent on $2$ is $3$ .

Worked example

Solve a logarithm equal to a number

Solve

\log_3(x-1)=2

.

Set the domain condition
The argument is $x-1$ . It must be positive, so any solution must be greater than $1$ .
$x-1>0\quad\Rightarrow\quad x>1$
Rewrite in exponential form
The logarithm asks which exponent on base $3$ gives the argument. Since the logarithm equals $2$ , the argument must equal $3$ to the power of $2$ .
$3^2=x-1$
Solve for the unknown
Evaluate the power, then add $1$ to both sides to isolate $x$ .
$9=x-1\quad\Rightarrow\quad x=10$
Check the answer
The value $10$ meets the restriction $x>1$ . Substitution gives an argument of $9$ , and $\log_3 9=2$ because $3^2=9$ .
$\log_3(10-1)=\log_3 9=2$

Answer:

x=10

Check: The argument is positive, and substituting

x=10

makes the original equation true.

Common mistakes and how to avoid them

Changing

\log_b A=c

into

A^b=c

.

Correction: The base is raised to the logarithm’s value. Rewrite it as

b^c=A

.

Accepting an answer without checking the argument.

Correction: Every logarithm’s argument must be positive. Check this condition and substitute into the original equation.

Treating the base and argument as interchangeable.

Correction: In

\log_b A

,

b

is the base and

A

is the argument. The argument is the value produced by raising the base to the exponent.

Lesson summary

A logarithm gives the exponent needed to produce its argument.
Use $\log_b A=c\iff b^c=A$ to solve a simple logarithmic equation.
For equal logarithms with the same base, equate their arguments, while requiring both arguments to be positive.
Check every proposed solution in the original equation.

Check your understanding

Question 1

Solve

\log_2(x+3)=4

.

$x=13$
$x=16$
$x=19$
$x=5$

Show answer and explanation

x=13

Rewrite as

2^4=x+3

. Then

16=x+3

, so

x=13

. The argument becomes

16

, which is positive.

Question 2

What condition must be true for

\log_5(3x-6)

to be defined?

$x>2$
$x<2$
$x\geq 2$
$x>0$

Show answer and explanation

x>2

The argument must be positive:

3x-6>0

. Dividing by

3

gives

x>2

.

Question 3

Which exponential equation is equivalent to

\log_4 64=3

?

$4^3=64$
$3^4=64$
$64^4=3$
$4^{64}=3$

Show answer and explanation

4^3=64

The base is

4

, the exponent is

3

, and the argument is

64

, so the equivalent statement is

4^3=64

.

Key terms

Logarithm: The exponent that tells how many times a base must be used as a power to produce a given positive value.
Base: The number raised to a power in an exponential expression or used as the base of a logarithm.
Argument: The expression inside a logarithm. It must have a positive value.
Domain condition: A restriction that tells which input values are allowed in an expression or equation.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation A3.3. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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