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A1.3 · Connect logarithmic and exponential equations

Learn to connect logarithmic and exponential equations through clear examples and targeted practice.

Ontario Grade 12 Mathematics

Exponential and Logarithmic Functions

Reading each form as a statement about the same exponent

A logarithm answers a question about an exponent. For example, asking “What exponent on 2 gives 8?” can be written as either 23=82^3=8 or log⁡28=3\log_2 8=3. These are not two different calculations. They are two ways to record the same relationship. In this lesson, you will connect the forms, use the connection to solve an equation, and check that the result makes sense.

What you will learn

1. Prerequisite bridge: powers and bases

An exponent tells how many times a base is used as a factor. In 32=93^2=9, the base is 33 and the exponent is 22. The equation says that raising 33 to the power 22 gives 99.
A logarithm reverses this question. The expression log⁡39\log_3 9 asks: “What exponent on 33 gives 99?” Since 32=93^2=9, the answer is 22. Thus, log⁡39=2\log_3 9=2.
In log⁡ba=c\log_b a=c, the base is bb, the argument is aa, and the logarithm's value is cc. The argument is the number inside the logarithm. For real logarithms, the base must be positive and not equal to 11, and the argument must be positive. These conditions matter when solving equations.
log⁡ba=c  ⟺  bc=a\log_b a=c\iff b^c=a

2. The connection in words, numbers, and symbols

The symbol   ⟺  \iff means “if and only if.” Here it marks two statements that mean exactly the same thing, as long as the base and argument meet the required conditions. To change from logarithmic form to exponential form, the base stays the base, the logarithm's value becomes the exponent, and the argument becomes the result.
Consider log⁡5125=3\log_5 125=3. The base is 55, the value is 33, and the argument is 125125. The matching power statement is 53=1255^3=125. In the other direction, 43=644^3=64 can be written as log⁡464=3\log_4 64=3.
This connection applies when the unknown is in different places. In log⁡2x=5\log_2 x=5, the logarithm asks which power of 22 equals xx. In 3x=203^x=20, the unknown is the exponent, so the logarithmic form is x=log⁡320x=\log_3 20. The second equation does not claim that xx is a whole number; the logarithm names the exact exponent.
log⁡ba=c  ⟺  bc=a\log_b a=c\iff b^c=a

3. Solving equations and checking restrictions

When a logarithmic equation has one logarithm equal to a number, rewrite it as a power equation. Then solve the resulting equation using familiar algebra. When an exponential equation has a variable exponent, rewrite it as a logarithmic equation if that gives a direct expression for the unknown.
For example, changing log⁡2(x+3)=4\log_2(x+3)=4 to exponential form gives 24=x+32^4=x+3. This turns the logarithm into an ordinary equation. Solve for xx, then check the original logarithm's argument. Since x+3x+3 must be positive, a proposed solution that makes x+3x+3 zero or negative is not allowed.
A check has two parts. First, substitute the candidate into the original equation and confirm the equality. Second, confirm that each logarithm has a positive argument and a valid base. A value that solves a rearranged equation is not a solution if it makes the original logarithm undefined.
Keep the order of the parts clear. In log⁡ba=c\log_b a=c, the base bb is the base of the power, while aa is the result. Reversing these roles changes the equation. For instance, log⁡28=3\log_2 8=3 corresponds to 23=82^3=8, not 82=38^2=3.
log⁡b(a)=c  ⟺  bc=a\log_b(a)=c\iff b^c=a

4. Applying the connection

The two forms are useful for different questions. Exponential form makes it easy to calculate a result when the exponent is known. Logarithmic form names the exponent when the result is known. For example, 25=322^5=32 quickly gives the value of a power, while log⁡232=5\log_2 32=5 directly states the exponent.
When solving, choose the form that exposes the unknown. If the unknown is inside the logarithm as part of its argument, converting to exponential form often produces an equation that can be solved with algebra. If the unknown is the exponent, logarithmic form can state its exact value.
The connection is also a way to check whether a proposed rewrite is correct. Read the logarithmic equation as a question: “The base raised to what exponent equals the argument?” If the power statement answers that question, the conversion is consistent.
bc=a  ⟺  log⁡ba=cb^c=a\iff \log_b a=c

Two forms, one relationship

Logarithmic formQuestion it answersMatching exponential form
log⁡5125=3\log_5 125=3What power of 55 gives 125125?53=1255^3=125
log⁡464=3\log_4 64=3What power of 44 gives 6464?43=644^3=64
log⁡320=x\log_3 20=xWhat power of 33 gives 2020?3x=203^x=20

Worked example

Solve a logarithmic equation by converting forms

Solve log⁡2(x+3)=4\log_2(x+3)=4 and check the result.
  1. Identify the parts
    The base is 22, the logarithm's value is 44, and its argument is x+3x+3. Rewrite the statement so that raising 22 to the power 44 gives the argument.
    24=x+32^4=x+3
  2. Evaluate the power
    Since 24=162^4=16, the equation states that x+3x+3 equals 1616. This is now a linear equation.
    16=x+316=x+3
  3. Isolate the variable
    Subtract 33 from both sides to leave xx by itself.
    x=13x=13
  4. Check the original equation
    The argument is 13+3=1613+3=16, which is positive. Also, 24=162^4=16, so the logarithm of 1616 with base 22 is 44. The candidate satisfies the original equation and its domain restriction.
    log⁡2(13+3)=log⁡216=4\log_2(13+3)=\log_2 16=4
Answer: x=13x=13
Check: The original logarithm is defined because its argument is 16>016>0, and its value is 44.

Common mistakes and how to avoid them

Writing the argument as the base when changing forms.
Correction: Keep the base unchanged. In log⁡ba=c\log_b a=c, the matching power is bc=ab^c=a.
Treating the logarithm's value as the result of the power.
Correction: The logarithm's value becomes the exponent. The argument becomes the result.
Keeping a candidate even though it makes a logarithm's argument zero or negative.
Correction: Check the original equation. Every logarithm's argument must be positive.
Assuming the unknown exponent must be a whole number.
Correction: An exponent can be a non-integer. A logarithm can represent the exact value of that exponent.

Lesson summary

Check your understanding

Question 1

Which exponential equation matches log⁡381=4\log_3 81=4?
  1. 34=813^4=81
  2. 43=814^3=81
  3. 381=43^{81}=4
  4. 813=481^3=4
Show answer and explanation
34=813^4=81
The base stays 33, the logarithm's value becomes the exponent 44, and the argument 8181 is the result.

Question 2

Which logarithmic equation matches 5x=255^x=25?
  1. log⁡525=x\log_5 25=x
  2. log⁡255=x\log_{25} 5=x
  3. log⁡5x=25\log_5 x=25
  4. log⁡x5=25\log_x 5=25
Show answer and explanation
log⁡525=x\log_5 25=x
The base is 55, the result is 2525, and the unknown exponent is the logarithm's value.

Question 3

Solve log⁡4(x−1)=2\log_4(x-1)=2. Which value is valid?
  1. x=17x=17
  2. x=5x=5
  3. x=0x=0
  4. x=15x=15
Show answer and explanation
x=17x=17
Convert to 42=x−14^2=x-1, so 16=x−116=x-1 and x=17x=17. Its argument is 17−1=16>017-1=16>0, and log⁡416=2\log_4 16=2.

Key terms

Base
The number being raised to a power; in log⁡ba\log_b a, it is bb.
Exponent
The number that tells which power of the base is used.
Argument
The number or expression inside a logarithm; in log⁡ba\log_b a, it is aa.
Logarithm
The exponent that makes a given base equal a specified result.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation A1.3. It is a study resource, not an official curriculum publication.

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