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A3.3 · Solve simple logarithmic equations

Learn to solve simple logarithmic equations through clear examples and targeted practice.

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⁡28\log_2 8 is the same as asking, “What power of 22 equals 88?” Since 23=82^3=8, the logarithm is 33. In this lesson, you will use that connection to solve simple equations and make sure each answer is allowed.

What you will learn

1. Prerequisite bridge: powers and logarithms

An exponent tells how many times a base is used as a factor. In 23=82^3=8, the base is 22, the exponent is 33, and the value is 88. A logarithm reverses this question: it gives the exponent when you know the base and the value.
In log⁡ba=c\log_b a=c, the base is bb, the argument is aa, and the logarithm’s value is cc. This statement means that bb raised to the power cc equals aa. The base must be positive and cannot equal 11, and the argument must be positive.
These conditions matter when solving equations. A logarithm such as log⁡5(x−2)\log_5(x-2) is defined only when its argument, x−2x-2, is positive. So its input must satisfy x>2x>2. This restriction is called a domain condition: it tells which input values are allowed.
log⁡ba=c  ⟺  bc=a\log_b a=c\iff b^c=a

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⁡bA=c\log_b A=c becomes bc=Ab^c=A, where AA is the expression inside the logarithm.
Before solving, note the domain condition A>0A>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⁡bA=log⁡bB\log_b A=\log_b B can be changed to A=BA=B, while keeping both conditions A>0A>0 and B>0B>0. Solve the resulting equation and test the domain conditions.
log⁡bA=log⁡bB  ⟺  A=B(A>0, B>0)\log_b A=\log_b B\iff A=B\quad(A>0,\ B>0)

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

Consider the question, “To what power must 33 be raised to make 2727?” The answer is 33, because 33=273^3=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⁡327=3  ⟺  33=27\log_3 27=3\iff 3^3=27

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+12x+1, require 2x+1>02x+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.
argument>0\text{argument}>0

One relationship in two forms

Logarithmic questionExponential statementMeaning
log⁡327=3\log_3 27=333=273^3=27The exponent on 33 is 33.
log⁡28=3\log_2 8=323=82^3=8The exponent on 22 is 33.

Worked example

Solve a logarithm equal to a number

Solve log⁡3(x−1)=2\log_3(x-1)=2.
  1. Set the domain condition
    The argument is x−1x-1. It must be positive, so any solution must be greater than 11.
    x−1>0⇒x>1x-1>0\quad\Rightarrow\quad x>1
  2. Rewrite in exponential form
    The logarithm asks which exponent on base 33 gives the argument. Since the logarithm equals 22, the argument must equal 33 to the power of 22.
    32=x−13^2=x-1
  3. Solve for the unknown
    Evaluate the power, then add 11 to both sides to isolate xx.
    9=x−1⇒x=109=x-1\quad\Rightarrow\quad x=10
  4. Check the answer
    The value 1010 meets the restriction x>1x>1. Substitution gives an argument of 99, and log⁡39=2\log_3 9=2 because 32=93^2=9.
    log⁡3(10−1)=log⁡39=2\log_3(10-1)=\log_3 9=2
Answer: x=10x=10
Check: The argument is positive, and substituting x=10x=10 makes the original equation true.

Common mistakes and how to avoid them

Changing log⁡bA=c\log_b A=c into Ab=cA^b=c.
Correction: The base is raised to the logarithm’s value. Rewrite it as bc=Ab^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⁡bA\log_b A, bb is the base and AA is the argument. The argument is the value produced by raising the base to the exponent.

Lesson summary

Check your understanding

Question 1

Solve log⁡2(x+3)=4\log_2(x+3)=4.
  1. x=13x=13
  2. x=16x=16
  3. x=19x=19
  4. x=5x=5
Show answer and explanation
x=13x=13
Rewrite as 24=x+32^4=x+3. Then 16=x+316=x+3, so x=13x=13. The argument becomes 1616, which is positive.

Question 2

What condition must be true for log⁡5(3x−6)\log_5(3x-6) to be defined?
  1. x>2x>2
  2. x<2x<2
  3. x≥2x\geq 2
  4. x>0x>0
Show answer and explanation
x>2x>2
The argument must be positive: 3x−6>03x-6>0. Dividing by 33 gives x>2x>2.

Question 3

Which exponential equation is equivalent to log⁡464=3\log_4 64=3?
  1. 43=644^3=64
  2. 34=643^4=64
  3. 644=364^4=3
  4. 464=34^{64}=3
Show answer and explanation
43=644^3=64
The base is 44, the exponent is 33, and the argument is 6464, so the equivalent statement is 43=644^3=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

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

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