1. Properties of Logarithms
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Chapter 5
1.
Properties of Logarithms
This lesson delves into the intriguing world of logarithms, highlighting their inverse relationship with exponents. Through various examples and explanations, readers are introduced to the properties of logarithms and their applications. Paulina, a character in the content, embarks on a journey to understand and master logarithmic expressions. She learns to rewrite exponential equations as logarithmic ones and vice versa. As she progresses, Paulina discovers how to evaluate, expand, and condense logarithmic expressions, both algebraic and numeric. By the end, she feels confident in her understanding and application of logarithmic properties.
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| | 11 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Properties of Logarithms
Slide of 11
Inverse operations are mathematical operations that, in a way, undo each other. This lesson will discuss logarithims, which are the inverse operation of raising a number to a variable. Properties of logarithms will also be described.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Explore
What Is the Inverse Operation of Calculating an Exponent?
Sometimes the inverse of a mathematical operation is clearly identifiable. For example, addition and subtraction, multiplication and division, or raising to the power of n and calculating the nth root are inverse operations.
However, in some other cases, inverse operations are a bit more difficult to come up with. What is an operation that undoes calculating an exponent? 
Discussion
Logarithm
A logarithm is the inverse function of an exponential function. The logarithm of a positive number m is written as logbm and read as the logarithm of m with base b.
logbm=n⇔bn=m
log416=n
In this equation, the definition of logarithm implies that n is the exponent to which the base 4 must be raised in order to obtain 16.
log416=n⇔4n=16
The following diagram illustrates how a logarithmic form has an equivalent exponential form using example numerical values.
Why
Logarithms are only defined for positive valuesLogarithms are undefined for non-positive values. The reason behind this can be explained by rewriting the logarithm in exponential form.
y=logbx⇔by=x
Because b is a positive number, the expression by is always positive. Therefore, since by=x, the value of x is always positive. This means that x can never be negative. In this case, x is the expression in the logarithm. Therefore, expressions in logarithmic functions must be positive.
As a consequence of the definition of a logarithm, two properties can be deduced. In these properties, b is positive and not equal to 1.
| Property | Reason |
|---|---|
| logbb=1 | A number raised to the power of 1 is equal to itself. |
| logb1=0 | A number raised to the power of 0 is equal to 1. |
Example
Evaluating and Rewriting Logarithmic Expressions
Paulina has recently become excited learning about logarithms.
She eagerly went to her math teacher and asked for some introductory exercises to practice evaluating and rewriting logarithmic expressions. Help her get off to a good start!
a Evaluate the expression log5125.
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b Rewrite the logarithmic equation log4a=x as an exponential equation.
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c Rewrite the exponential equation 3x=a as a logarithmic equation.
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Hint
a To what power must the base of 5 be raised to obtain 125?
b Use the definition of a logarithm.
c How can an exponential equation be rewritten using a logarithm?
Solution
a To evaluate the given expression, it is helpful to recall what the definition of a logarithm is.
logba=c⇔bc=a
With this definition in mind, let x be the value of log5125. log5125=x⇔5x=125
Given that 5 is the base, Paulina should ask herself what number it must be raised to in order to reach 125. Well, 5 to the power of 3 is equal to 125, the value of x is 3. Therefore, log5125=3.
log5125=3⇔53=125
b Similar to Part A, to rewrite log4a=x as an exponential equation, the definition of a logarithm will be used.
logba=c⇔bc=a
Therefore, Paulina should substitute b=4 and c=x in the above definition.
log4a=x⇔4x=a
c To rewrite 3x=a as a logarithmic equation, the definition of a logarithm will be used one last time.
bc=a⇔logba=c
Here, b=3 and c=x will be substituted into the definition.
3x=a⇔log3a=x
Pop Quiz
Evaluating Logarithms
Evaluate the logarithms.
Pop Quiz
Rewriting Logarithmic and Exponential Equations
Rewrite the logarithmic equations as exponential equations and the exponential equations as logarithmic equations.
Discussion
Properties of Logarithms
For the proof of these properties, two identities will be used. Start by recalling the definition of a logarithm.
The obtained identities will be used together with the Product of Powers Property to prove the Product Property of Logarithms.
The obtained identities will be used together with the Quotient of Powers Property to prove the Quotient Property of Logarithms.
The obtained identities will be used together with the Power of a Power Property to prove the Power Property of Logarithms.
logba=c ⇔ a=bc
The first equation of the definition states that c=logba. Therefore logba can be substituted for c in the second equation. Furthermore, the second equation states that a is equal to bc. This means that bc can be substituted for a in the first equation. | logba=c ⇔ a=bc |
|---|
| a=bc Substitute a=blogba |
| logba=c Substitute logbbc=c |
With this information in mind, three properties can be stated.
Rule
Product Property of Logarithms
The logarithm of a product can be written as the sum of the individual logarithms of each factor.
logbmn=logbm+logbn
log3(7⋅4)=log37+log34
Proof
The obtained identities will be used together with the Product of Powers Property to prove the Product Property of Logarithms.
logbmn
Rewrite
Rewrite mn as m⋅n
logb(m⋅n)
m=blogb(m)
logb(blogbm⋅blogbn)
MultPow
am⋅an=am+n
logb(blogbm+logbn)
logb(bm)=m
logbm+logbn
Rule
Quotient Property of Logarithms
The logarithm of a quotient can be written as the difference between the logarithm of the numerator and the logarithm of the denominator.
logbnm=logbm−logbn
log347=log37−log34
Proof
The obtained identities will be used together with the Quotient of Powers Property to prove the Quotient Property of Logarithms.
logbnm
m=blogb(m)
logb(blogbnblogbm)
DivPow
anam=am−n
logb(blogbm−logbn)
logb(bm)=m
logbm−logbn
Rule
Power Property of Logarithms
The logarithm of a power can be written as the product of the exponent and the logarithm of the base.
logbmn=nlogbm
log274=4log27
Proof
The obtained identities will be used together with the Power of a Power Property to prove the Power Property of Logarithms.
logbmn
m=blogb(m)
logb(blogbm)n
PowPow
(am)n=am⋅n
logbb(logbm)⋅n
logb(bm)=m
(logbm)⋅n
CommutativePropMult
Commutative Property of Multiplication
nlogbm
Example
Using Properties to Evaluate Expressions
After understanding the definition of a logarithm and learning about its properties, Paulina is ready to delve deeper into this topic.
As a part of her deep dive into properties of logarithms, she begins with two approximations.
log23≈1.585andlog25≈2.322
Without using a calculator, only relying on the two given approximations, Paulina wants to evaluate three logarithmic expressions. This will make her feel like she is really understaning how to operate with logarithms. Help her find the answer! Write each answer rounded to three decimal places. a log215
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b log251
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c log224
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Hint
a Use the Product Property of Logarithms.
b Use the Quotient Property of Logarithms.
c Use the Power Property of Logarithms.
Solution
a Consider the Product Property of Logarithms.
logbmn=logbm+logbn
Knowing that log23≈1.585 and that log25≈2.322, this property can be used to evaluate the given expression. Start by rewriting 15 as the product of 3 and 5.
log26
SplitIntoFactors
Split into factors
log2(3⋅5)
log2(mn)=log2(m)+log2(n)
log23+log25
log23≈1.585, log25≈2.322
1.585+2.322
AddTerms
Add terms
3.907
b Because of the quotient inside the logarithm, the Quotient Property of Logarithms will be used this time.
logbnm=logbm−logbn
Also, as a consequence of the definition of a logarithm, it is known that logb1=0 for any positive number b different than 1. Using this information, the above property, and the fact that log25≈2.322, the given expression can be evaluated.
log251
log2(ba)=log2(a)−log2(b)
log21−log25
log2(1)=0
0−log25
log25≈2.322
0−2.322
SubTerm
Subtract term
-2.322
c Recall the Power Property of Logarithms and the Product Property of Logarithms.
Power: Product: logbmn=nlogbm logbmn=logbm+logbn
Also, as a consequence of the definition of a logarithm, it is known that logbb=1 for any positive number b different than 1. Using this information, the above properties, and the fact that log23≈1.585, the given expression can be evaluated.
log224
SplitIntoFactors
Split into factors
log2(8⋅3)
log2(mn)=log2(m)+log2(n)
log28+log23
WritePow
Write as a power
log223+log23
log2(am)=m⋅log2(a)
3log22+log23
log2(2)=1
3(1)+log23
IdPropMult
Identity Property of Multiplication
3+log23
log23≈1.585
3+1.585
AddTerms
Add terms
4.585
Example
Using Properties to Expand Expressions
Paulina has moved beyond only understanding the definition of a logarithm. She can now rewrite exponential equations as logarithmic equations and convert logarithmic equations into exponential equations. On top of all of that, she can even evaluate logarithmic expressions. Paulina is starting to master this topic! 
Now, there are even more interesting challenges to face. This time, Paulina needs to expand two logarithmic expressions by using the Properties of Logarithms. Help her do this! Assume that all the variables involved are positive.
a log5y225x3
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b log44xy
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Hint
a Start by using the Quotient Property of Logarithms. Then, use the Product Property of Logarithms. Finally, use the Power Property of Logarithms.
b Start by rewriting the square root as a rational exponent. Then, use the Power Property of Logarithms to remove the exponent.
Solution
a Consider the Properties of Logarithms.
Product:Quotient:Power: logbmn=logbm+logbn logbnm=logbm−logbn logbmn=nlogbm
Here, b, m, and n are positive, where b=1. The main operation in the given logarithmic expression is a division. Therefore, the first property that should be used is the Quotient Property of Logarithms. Then, the Product Property of Logarithms and the Power Property of Logarithms will be applied.
log5y225x3
log5(ba)=log5(a)−log5(b)
log525x3−log5y2
log5(mn)=log5(m)+log5(n)
log525+log5x3−log5y2
log5(am)=m⋅log5(a)
log525+3log5x−2log5y
Calculate logarithm
2+3log5x−2log5y
b To use the Power Property of Logarithms, the square root will be written as a rational exponent.
log44xy=log4(4xy)21
Now, the mentioned property can be used. Then, the Quotient Property of Logarithms and the Product Property of Logarithms can also be used.
log4(4xy)21
log4(am)=m⋅log4(a)
21log44xy
log4(ba)=log4(a)−log4(b)
21(log4xy−log44)
log4(mn)=log4(m)+log4(n)
21(log4x+log4y−log44)
Distr
Distribute 21
21log4x+21log4y−21log44
log4(4)=1
21log4x+21log4y−21(1)
IdPropMult
Identity Property of Multiplication
21log4x+21log4y−21
Example
Using Properties to Condense Expressions
Paulina feels extremely confident about her skills in using logarithms! Now, instead of expanding algebraic and numeric expressions that involve logarithms, she will practice condensing them.
To finally master logarithmic expressions with and without variables, Paulina wants to condense two expressions using the Properties of Logarithms. Help her do this! Assume that all variables involved are positive.
a log212+3log25−log26
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b 5log32−6log3x+2log3y
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Hint
a Use the Product Property of Logarithms, the Power Property of Logarithms, and the Quotient Property of Logarithms.
b Use the Properties of Logarithms.
Solution
a To condense this expression, three properties of logarithms will be recalled first. These are the Product Property of Logarithms, the Quotient Property of Logarithms, and the Power Property of Logarithms.
Product:Quotient:Power: logbmn=logbm+logbn logbnm=logbm−logbn logbmn=nlogbm
These three properties will be used to condense the expression. log212+3log25−log26
m⋅log2(a)=log2(am)
log212+log253−log26
CalcPow
Calculate power
log212+log2125−log26
log2(m)+log2(n)=log2(mn)
log2(12⋅125)−log26
Multiply
Multiply
log21500−log26
log2(m)−log2(n)=log2(nm)
log261500
CalcQuot
Calculate quotient
log2250
b Similar to Part A, to condense this expression, the same three Properties of Logarithms will be applied.
5log32−6log3x+2log3y
m⋅log3(a)=log3(am)
log325−log3x6+log3y2
CalcPow
Calculate power
log332−log3x6+log3y2
log3(m)−log3(n)=log3(nm)
log3x632+log3y2
log3(m)+log3(n)=log3(mn)
log3(x632⋅y2)
MoveRightFacToNum
ca⋅b=ca⋅b
log3x632y2
Closure
Inverse Operations
In this lesson, it has been presented that exponents and logarithms are their own inverses. This means that these two operations essentially undo each other.
The definition of a logarithm and important Properties of Logarithms have also been taught and practiced in this lesson.
| Definition | logba=c ⇔ a=bc |
|---|---|
| Identity Derived From the Definition | logbb=1 |
| Identity Derived From the Definition | logb1=0 |
| Product Property of Logarithms | logbmn=logbm+logbn |
| Quotient Property of Logarithms | logbnm=logbm−logbn |
| Power Property of Logarithms | logbmn=nlogbm |
It is important to keep in mind that these properties are only valid for positive values of a, b, m, and n, where b=1. Furthermore, these properties can be used in several situations.
- Writing a power as a logarithm and writing a logarithm as a power.
- Evaluating logarithms.
- Using given approximations of certain logarithms to approximate other logarithmic expressions.
- Expanding and condensing logarithmic expressions with and without variables.
Properties of Logarithms
Exercises
Evaluate the logarithms.
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Let's begin by recalling the definition of a logarithm. log_b a = c ⇔ b^c=a In order to solve the given logarithmic expression, we need to think about which number we need to raise 3 to in order to get 81. log_3 81=? ⇔ 3^? =81 To find the number we are looking for, we will make a table of values to calculate the value of different powers of 3.
| Powers of 3 | |
|---|---|
| Exponent | Power |
| 0 | 3^0=1 |
| 1 | 3^1=3 |
| 2 | 3^2=9 |
| 3 | 3^3=27 |
| 4 | 3^4=81 |
Because 3 raised to the fourth power equals 81, we know that log_3 81 is equal to 4. 3^4 = 81 ⇔ log_3 81= 4
Since any number raised to the power of 1 is equal to itself, we know that 3 raised to the power of 1 equals 3. We can use this information and the definition of a logarithm to find the value of log_3 3. 3^1= 3 ⇔ log_3 3= 1
Any non-zero number raised to the power of 0 is equal to 1. As such, we know that 12 raised to the power of 0 equals 1. We can use this information and the definition of a logarithm to find the value of log_(12) 1. ( 1/2)^0= 1 ⇔ log_(12) 1= 0
Let's start by rewriting 0.25 as 14. log_4 0.25 ⇔ log_4 1/4 Next, recall the Negative Exponent Property. a^(- n)=1/a^n, fora≠ 0andnnatural Therefore, 4 raised to the power of - 1 is equal to 14^1, or 14. By using this information and the definition of a logarithm, we know that the value of log_4 14 is - 1. 4^(- 1)= 1/4 ⇔ log_4 1/4= - 1 This means that log_4 0.25=- 1.
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Rewrite the logarithmic expressions as exponential expressions.
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Let's start by recalling the definition of a logarithm. log_b a= c ⇔ b^c= a We can use this definition to rewrite the given logarithmic expression as an exponential expression. log_5 x= y ⇔ 5^y= x
As we did in Part A, let's start by recalling the definition of a logarithm.
log_b a= c ⇔ b^c= a
Now let's rewrite the given logarithmic expression as an exponential expression.
log_b 10= y ⇔ b^y= 10
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Rewrite the exponential expressions as logarithmic expressions.
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Let's start by recalling the definition of a logarithm. b^c= a ⇔ log_b a= c We can use this definition to rewrite the given exponential expression as a logarithmic expression. 3^m= n ⇔ log_3 n= m
Like we did in part A, let's start by recalling the definition of a logarithm.
b^c= a ⇔ log_b a= c
Now let's rewrite the given exponential expression as a logarithmic expression.
x^2= y ⇔ log_x y= 2
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log74 log712 ≈0.712≈1.277
Use these approximations and the Properties of Logarithms to find the numerical value of the following logarithmic expressions.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.48312em;vertical-align:0em;\"><\/span><span class=\"mrel\">\u2248<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["1.424"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.48312em;vertical-align:0em;\"><\/span><span class=\"mrel\">\u2248<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["-0.712"]}}
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We can use the fact that log_7 4 ≈ 0.712 to approximate the value of the given logarithmic expression. To expand the expression, first recall the Power Property of Logarithms. log_b m^n = n log_b m In this formula, m, n, and b are positive numbers, where b≠ 1. We can now rewrite 16 as a power of 4, apply the property, and then substitute 0.712 for log_7 4.
We can use the fact that log_7 4 ≈ 0.712 again to approximate the value of the given logarithmic expression. To expand the expression, we will first recall the Quotient Property of Logarithms. log_b m/n=log_b m - log_b n In this formula, m, n, and b are positive numbers, where b≠ 1. Let's apply the property and then substitute 0.712 for log_7 4. Keep in mind that log_b 1=0.
We can use the given values to approximate the value of the given logarithmic expression. To expand the expression, we will first recall the Quotient Property of Logarithms. log_b m/n=log_b m - log_b n In this formula, m, n, and b are positive numbers, where b≠ 1. Next, we can find an equivalent fraction to 13 with denominator 12, apply the property, and then substitute 0.712 for log_7 4 and 1.277 for log_7 12.
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Use the properties of logarithms to fully expand the logarithmic expressions.
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x","y"],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["\\ln 6+2\\ln x-4\\ln y","\\ln 6-4\\ln y+2\\ln x","2\\ln x-4\\ln y+\\ln 6","2\\ln x+\\ln 6-4\\ln y","-4\\ln y+\\ln 6+2\\ln x","-4\\ln y+2\\ln x+\\ln 6"]}}
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To expand the given logarithmic expression, we will first use the Quotient Property of Logarithms and then the Product Property of Logarithms.
To expand the given logarithmic expression, we will use the Product Property of Logarithms, the Quotient Property of Logarithms, and then the Power Property of Logarithms.
To expand the given logarithmic expression, we will start by rewriting sqrt(x) as x^(12). Then, we will use the Product Property of Logarithms and the Power Property of Logarithms. Let's do it!
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Use the properties of logarithms to fully condense the logarithmic expressions.
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To write the given expression as a single logarithm, we will use the Power Property of Logarithms and the Quotient Property of Logarithms. Let's do it!
To write the given expression as a single logarithm, we will use the Power Property of Logarithms and the Product Property of Logarithms. Let's do it!
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Properties of Logarithms
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