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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Lesson Settings & Tools
| | 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.
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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 n^\text{th} 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? 
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Logarithm
A logarithm is the inverse function of an exponential function. The logarithm of a positive number m is written as log_b m and read as the logarithm of m with base b.
log_b m=n ⇔ b^n=m
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=log_b x ⇔ b^y=x Because b is a positive number, the expression b^y is always positive. Therefore, since b^y=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 |
|---|---|
| log_b b=1 | A number raised to the power of 1 is equal to itself. |
| log_b 1=0 | A number raised to the power of 0 is equal to 1. |
Example
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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 log_5 125.
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b Rewrite the logarithmic equation log_4 a=x as an exponential equation.
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c Rewrite the exponential equation 3^x=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.
log_b a= c ⇔ b^c= a With this definition in mind, let x be the value of log_5 125. log_5 125= x ⇔ 5^x= 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, log_5 125=3. log_5 125= 3 ⇔ 5^3= 125
b Similar to Part A, to rewrite log_4 a=x as an exponential equation, the definition of a logarithm will be used.
log_b a= c ⇔ b^c= a Therefore, Paulina should substitute b= 4 and c= x in the above definition. log_4 a= x ⇔ 4^x= a
c To rewrite 3^x=a as a logarithmic equation, the definition of a logarithm will be used one last time.
b^c= a ⇔ log_b a= c Here, b= 3 and c= x will be substituted into the definition. 3^x= a ⇔ log_3 a= x
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Evaluating Logarithms
Evaluate the logarithms.
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Rewriting Logarithmic and Exponential Equations
Rewrite the logarithmic equations as exponential equations and the exponential equations as logarithmic equations.
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Properties of Logarithms
For the proof of these properties, two identities will be used. Start by recalling the definition of a logarithm. log_b a=c ⇔ a=b^c The first equation of the definition states that c=log_b a. Therefore log_b a can be substituted for c in the second equation. Furthermore, the second equation states that a is equal to b^c. This means that b^c can be substituted for a in the first equation.
| log_b a=c ⇔ a=b^c |
|---|
| a=b^c Substitute a=b^(log_b a) |
| log_b a=c Substitute log_b b^c=c |
With this information in mind, three properties can be stated.
Rule
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Product Property of Logarithms
The logarithm of a product can be written as the sum of the individual logarithms of each factor.
log_b mn=log_b m+log_b n
This property is only valid for positive values of b, m, and n, and for b≠ 1. As an example, the expression log_3 (7*4) can be rewritten using this property. log_3 (7*4)=log_3 7+log_3 4
Proof
The obtained identities will be used together with the Product of Powers Property to prove the Product Property of Logarithms.
log_b mn
Rewrite
Rewrite mn as m* n
log_b(m* n)
m=b^(log_b(m))
log_b(b^(log_b m)* b^(log_b n))
MultPow
a^m*a^n=a^(m+n)
log_b(b^(log_b m+log_b n))
log_b(b^m)=m
log_b m+log_b n
Rule
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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.
log_b m/n=log_b m -log_b n
This property is valid for positive values of b, m, and n, and for b≠ 1. For example, the expression log_3 74 can be rewritten using this property. log_3 7/4=log_3 7-log_3 4
Proof
The obtained identities will be used together with the Quotient of Powers Property to prove the Quotient Property of Logarithms.
log_b m/n
m=b^(log_b(m))
log_b(b^(log_b m)/b^(log_b n))
DivPow
a^m/a^n= a^(m-n)
log_b(b^(log_b m-log_b n))
log_b(b^m)=m
log_b m-log_b n
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Power Property of Logarithms
The logarithm of a power can be written as the product of the exponent and the logarithm of the base.
log_b m^n =nlog_b m
This property is valid for positive values of b, m, and n, and for b≠ 1. For example, log_2 7^4 can be rewritten using this property. log_2 7^4=4 log_2 7
Proof
The obtained identities will be used together with the Power of a Power Property to prove the Power Property of Logarithms.
log_b m^n
m=b^(log_b(m))
log_b (b^(log_b m) )^n
PowPow
(a^m)^n=a^(m* n)
log_b b^((log_b m )* n)
log_b(b^m)=m
(log_b m)* n
CommutativePropMult
Commutative Property of Multiplication
nlog_b m
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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.
log_2 3≈ 1.585 and log_2 5≈ 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 log_2 15
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b log_2 1/5
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c log_2 24
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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.
log_2 6
SplitIntoFactors
Split into factors
log_2 (3* 5)
log_2(mn)=log_2(m) + log_2(n)
log_2 3+ log_2 5
log_2 3 ≈ 1.585, log_2 5 ≈ 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.
log_2 1/5
log_2(a/b)=log_2(a) - log_2(b)
log_2 1-log_2 5
log_2(1) = 0
0-log_2 5
log_2 5 ≈ 2.322
0- 2.322
SubTerm
Subtract term
-2.322
c Recall the Power Property of Logarithms and the Product Property of Logarithms.
log_2 24
SplitIntoFactors
Split into factors
log_2 (8* 3)
log_2(mn)=log_2(m) + log_2(n)
log_2 8 +log_2 3
WritePow
Write as a power
log_2 2^3 +log_2 3
log_2(a^m)= m* log_2(a)
3log_2 2 +log_2 3
log_2(2) = 1
3(1)+log_2 3
IdPropMult
Identity Property of Multiplication
3+log_2 3
log_2 3 ≈ 1.585
3+ 1.585
AddTerms
Add terms
4.585
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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 log_5 25x^3/y^2
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b log_4 sqrt(xy/4)
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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.
log_5 25x^3/y^2
log_5(a/b)=log_5(a) - log_5(b)
log_5 25x^3-log_5 y^2
log_5(mn)=log_5(m) + log_5(n)
log_5 25+log_5 x^3-log_5 y^2
log_5(a^m)= m* log_5(a)
log_5 25+3log_5 x-2log_5 y
Calculate logarithm
2+3log_5 x-2log_5 y
b To use the Power Property of Logarithms, the square root will be written as a rational exponent.
log_4 (xy/4)^(12)
log_4(a^m)= m* log_4(a)
1/2log_4 xy/4
log_4(a/b)=log_4(a) - log_4(b)
1/2(log_4 xy-log_4 4)
log_4(mn)=log_4(m) + log_4(n)
1/2(log_4 x+log_4 y-log_4 4)
Distr
Distribute 1/2
1/2 log_4 x+1/2 log_4 y-1/2log_4 4
log_4(4) = 1
1/2 log_4 x+1/2 log_4 y-1/2(1)
IdPropMult
Identity Property of Multiplication
1/2 log_4 x+1/2 log_4 y-1/2
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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 log_2 12 +3log_2 5-log_2 6
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b 5log_3 2-6log_3 x+2log_3 y
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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.
log_2 12 +3log_2 5-log_2 6
m* log_2(a)=log_2(a^m)
log_2 12 +log_2 5^3-log_2 6
CalcPow
Calculate power
log_2 12 +log_2 125-log_2 6
log_2(m) + log_2(n)=log_2(mn)
log_2 (12 * 125)-log_2 6
Multiply
Multiply
log_2 1500-log_2 6
log_2(m) - log_2(n)=log_2(m/n)
log_2 1500/6
CalcQuot
Calculate quotient
log_2 250
b Similar to Part A, to condense this expression, the same three Properties of Logarithms will be applied.
5log_3 2-6log_3 x+2log_3 y
m* log_3(a)=log_3(a^m)
log_3 2^5-log_3 x^6+log_3 y^2
CalcPow
Calculate power
log_3 32-log_3 x^6+log_3 y^2
log_3(m) - log_3(n)=log_3(m/n)
log_3 32/x^6+log_3 y^2
log_3(m) + log_3(n)=log_3(mn)
log_3 (32/x^6* y^2)
MoveRightFacToNum
a/c* b = a* b/c
log_3 32y^2/x^6
Closure
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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 | log_b a=c ⇔ a=b^c |
|---|---|
| Identity Derived From the Definition | log_b b=1 |
| Identity Derived From the Definition | log_b 1=0 |
| Product Property of Logarithms | log_b mn = log_b m+log_b n |
| Quotient Property of Logarithms | log_b m/n = log_b m-log_b n |
| Power Property of Logarithms | log_b m^n = nlog_b m |
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
Exercise 2.1
Order the logarithms from least to greatest value.
log_5 23, log_6 38, log_7 8, log_2 10
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To order the given logarithms from least to greatest value, we will find lower and upper boundaries for the numerical value of each expression. Let's start with log_5 23. Recall that 5^1=5 and that 5^2=25. With this information, we can find the values of log_5 5 and log_5 25.
cc
5^1= 5 & 5^2= 25
⇕ & ⇕
log_5 5 = 1 & log_5 25= 2
Because 23 is between 5 and 25, we can state that log_5 23 is between log_5 5 and log_5 25. This means that log_5 23 is greater than 1 and less than 2.
1
| Inequality | Comment |
|---|---|
| 1 | log_5 23 is closer to 2 than to 1 |
| 2 | - |
| 1 | log_7 8 is closer to 1 than to 2 |
| log_2 10>3 | - |
By using the information from the table, we can write the logarithmic expressions in order from least to greatest value. log_7 8, log_5 23, log_6 38, log_2 10
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Use the definition of a logarithm to evaluate the logarithms. Write the answers as simplified fractions.
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Let's start by writing 125 and 25, which are the base and the argument of the given logarithm, respectively, as powers with base 5. log_(125) 25 = log_(5^3) 5^2 Therefore, we need to find a number c such that 5^3 raised to the power of c equals 5^2. This can be algebraically expressed by using the definition of a logarithm. log_(5^3) 5^2= c ⇔ ( 5^3)^c= 5^2 We can use the Power of a Power Property to solve the exponential equation. Then, since both sides have the same base, we can use the Property of Equality for Exponential Equations and equate the exponents. Let's do it!
We found that the numerical value of log_(125) 25 is 23.
Similar to Part A, we can start by expressing the base of the logarithm 8 and the argument 32 as powers with base 2.
log_8 32 = log_(2^3) 2^5
We need to find a number c such that 2^3 raised to the power of c equals 2^5. Let's use the definition of a logarithm to algebraically express this.
log_(2^3) 2^5= c ⇔ ( 2^3)^c= 2^5
We can solve the exponential equation by using the Power of a Power Property and the Property of Equality for Exponential Equations.
Therefore, the numerical value of log_8 32 is 53.
In a similar manner as before, we can start by expressing the base of the logarithm 27 and the argument 81 as powers with base 3.
log_(27) 81 = log_(3^3) 3^4
We need to find a number c such that 3^3 raised to the power of c equals 3^4. Let's use the definition of a logarithm to algebraically express this.
log_(3^3) 3^4= c ⇔ ( 3^3)^c= 3^4
We can solve the exponential equation by using the Power of a Power Property and the Property of Equality for Exponential Equations.
Therefore, the numerical value of log_(27) 81 is 43.
Finally, we can express 4 and 128, which are the base and the argument of the given logarithm, as powers with base 2.
log_4 128 = log_(2^2) 2^7
We need to find a number c such that 2^2 raised to the power of c equals 2^7. Let's use the definition of a logarithm to algebraically express this.
log_(2^2) 2^7= c ⇔ ( 2^2)^c= 2^7
We can solve the exponential equation by using the Power of a Power Property and the Property of Equality for Exponential Equations.
We found that the numerical value of log_4 128 is 72.
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Consider the following logarithmic expression.
log x =5
What is the value of 1x?
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Before finding the value of 1x, it would be convenient for us to find the value of x. To do so, we will use the definition of a logarithm. Note that we have a common logarithm, which means that the base is 10. Definition:& log_b a= c ⇔ b^c= a Given Logarithm:& log x = 5 ⇔ 10^5= x We found that x is equal to 10^5, or 100 000. With this information, we can find the value of its reciprocal 1x. x=100 000 ⇔ 1/x=1/100 000 Therefore, 1x is equal to 1100 000 or 0.00001.
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Properties of Logarithms
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