{{ 'ml-label-loading-course' | message }}

{{ tocSubheader }}

{{ 'ml-toc-proceed-mlc' | message }}

{{ 'ml-toc-proceed-tbs' | message }}

An error ocurred, try again later!

Chapter {{ article.chapter.number }}

{{ article.number }}. # {{ article.displayTitle }}

{{ article.intro.summary }}

Show less Show more Lesson Settings & Tools

| {{ 'ml-lesson-number-slides' | message : article.intro.bblockCount }} |

| {{ 'ml-lesson-number-exercises' | message : article.intro.exerciseCount }} |

| {{ 'ml-lesson-time-estimation' | message }} |

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

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 $n_{th}$ root are inverse operations.
*undoes* calculating an exponent?

However, in some other cases, inverse operations are a bit more difficult to come up with. What is an operation that

Discussion

A logarithm is the inverse function of an exponential function. The logarithm of a positive number $m$ is written as $g_{b}m$ and read as the logarithm of $m$ with base $b.$

$g_{b}m=n⇔b_{n}=m$

$g_{4}16=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 $16.$
$g_{4}16=n ⇔4_{n}=16 $

The following diagram illustrates how a Logarithms are undefined for non-positive values. The reason behind this can be explained by rewriting the logarithm in exponential form.
*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.

$y=g_{b}x⇔b_{y}=x $

Because $b$ is a positive number, the expression $b_{y}$ is 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 |
---|---|

$g_{b}b=1$ | A number raised to the power of $1$ is equal to itself. |

$g_{b}1=0$ | A number raised to the power of $0$ is equal to $1.$ |

Example

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 $g_{5}125.$

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["3"]}}

b Rewrite the logarithmic equation $g_{4}a=x$ as an exponential equation.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":["a"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["a=4^x","4^x=a"]}}

c Rewrite the exponential equation $3_{x}=a$ as a logarithmic equation.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":["a"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["x=\\log_3a","\\log_3a=x","x=\\log_3(a)","\\log_3(a)=x"]}}

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?

a To evaluate the given expression, it is helpful to recall what the definition of a logarithm is.

$g_{b}a=c⇔b_{c}=a $

With this definition in mind, let $x$ be the value of $g_{5}125.$ $g_{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, $g_{5}125=3.$
$g_{5}125=3⇔5_{3}=125 $

b Similar to Part A, to rewrite $g_{4}a=x$ as an exponential equation, the definition of a logarithm will be used.

$g_{b}a=c⇔b_{c}=a $

Therefore, Paulina should substitute $b=4$ and $c=x$ in the above definition.
$g_{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⇔g_{b}a=c $

Here, $b=3$ and $c=x$ will be substituted into the definition.
$3_{x}=a⇔g_{3}a=x $

Pop Quiz

Evaluate the logarithms.

Pop Quiz

Rewrite the logarithmic equations as exponential equations and the exponential equations as logarithmic equations.

Discussion

For the proof of these properties, two identities will be used. Start by recalling the definition of a logarithm.

## Product Property of Logarithms

This property is only valid for positive values of $b,$ $m,$ and $n,$ and for $b =1.$ As an example, the expression $g_{3}(7⋅4)$ can be rewritten using this property.
### Proof

The obtained identities will be used together with the Product of Powers Property to prove the Product Property of Logarithms.
## Quotient Property of Logarithms

This property is valid for positive values of $b,$ $m,$ and $n,$ and for $b =1.$ For example, the expression $g_{3}47 $ can be rewritten using this property.
### Proof

The obtained identities will be used together with the Quotient of Powers Property to prove the Quotient Property of Logarithms.
## Power Property of Logarithms

This property is valid for positive values of $b,$ $m,$ and $n,$ and for $b =1.$ For example, $g_{2}7_{4}$ can be rewritten using this property.
### Proof

The obtained identities will be used together with the Power of a Power Property to prove the Power Property of Logarithms.

$g_{b}a=c⇔a=b_{c} $

The first equation of the definition states that $c=g_{b}a.$ Therefore $g_{b}a$ 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. $g_{b}a=c⇔a=b_{c}$ |
---|

$a=b_{c}Substitute a=b_{log_{b}a}$ |

$g_{b}a=cSubstitute g_{b}b_{c}=c$ |

With this information in mind, three properties can be stated.

Rule

The logarithm of a product can be written as the sum of the individual logarithms of each factor.

$g_{b}mn=g_{b}m+g_{b}n$

$g_{3}(7⋅4)=g_{3}7+g_{3}4 $

The obtained identities will be used together with the Product of Powers Property to prove the Product Property of Logarithms.

$g_{b}mn$

Rewrite

Rewrite $mn$ as $m⋅n$

$g_{b}(m⋅n)$

$m=b_{log_{b}(m)}$

$g_{b}(b_{log_{b}m}⋅b_{log_{b}n})$

MultPow

$a_{m}⋅a_{n}=a_{m+n}$

$g_{b}(b_{log_{b}m+log_{b}n})$

$g_{b}(b_{m})=m$

$g_{b}m+g_{b}n$

Rule

The logarithm of a quotient can be written as the difference between the logarithm of the numerator and the logarithm of the denominator.

$g_{b}nm =g_{b}m−g_{b}n$

$g_{3}47 =g_{3}7−g_{3}4 $

The obtained identities will be used together with the Quotient of Powers Property to prove the Quotient Property of Logarithms.

$g_{b}nm $

$m=b_{log_{b}(m)}$

$g_{b}(b_{log_{b}n}b_{log_{b}m} )$

DivPow

$a_{n}a_{m} =a_{m−n}$

$g_{b}(b_{log_{b}m−log_{b}n})$

$g_{b}(b_{m})=m$

$g_{b}m−g_{b}n$

Rule

The logarithm of a power can be written as the product of the exponent and the logarithm of the base.

$g_{b}m_{n}=ng_{b}m$

$g_{2}7_{4}=4g_{2}7 $

The obtained identities will be used together with the Power of a Power Property to prove the Power Property of Logarithms.

$g_{b}m_{n}$

$m=b_{log_{b}(m)}$

$g_{b}(b_{log_{b}m})_{n}$

PowPow

$(a_{m})_{n}=a_{m⋅n}$

$g_{b}b_{(log_{b}m)⋅n}$

$g_{b}(b_{m})=m$

$(g_{b}m)⋅n$

CommutativePropMult

Commutative Property of Multiplication

$ng_{b}m$

Example

After understanding the definition of a logarithm and learning about its properties, Paulina is ready to delve deeper into this topic.
### Hint

### Solution

As a part of her deep dive into properties of logarithms, she begins with two approximations.

$g_{2}3≈1.585andg_{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 $g_{2}15$

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["3.907"]}}

b $g_{2}51 $

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["- 2.322"]}}

c $g_{2}24$

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["4.585"]}}

a Use the Product Property of Logarithms.

b Use the Quotient Property of Logarithms.

c Use the Power Property of Logarithms.

a Consider the Product Property of Logarithms.

$g_{b}mn=g_{b}m+g_{b}n $

Knowing that $g_{2}3≈1.585$ and that $g_{2}5≈2.322,$ this property can be used to evaluate the given expression. Start by rewriting $15$ as the product of $3$ and $5.$
$g_{2}6$

SplitIntoFactors

Split into factors

$g_{2}(3⋅5)$

$g_{2}(mn)=g_{2}(m)+g_{2}(n)$

$g_{2}3+g_{2}5$

$g_{2}3≈1.585$, $g_{2}5≈2.322$

$1.585+2.322$

AddTerms

Add terms

$3.907$

b Because of the quotient inside the logarithm, the Quotient Property of Logarithms will be used this time.

$g_{b}nm =g_{b}m−g_{b}n $

Also, as a consequence of the definition of a logarithm, it is known that $g_{b}1=0$ for any positive number $b$ different than $1.$ Using this information, the above property, and the fact that $g_{2}5≈2.322,$ the given expression can be evaluated.
$g_{2}51 $

$g_{2}(ba )=g_{2}(a)−g_{2}(b)$

$g_{2}1−g_{2}5$

$g_{2}(1)=0$

$0−g_{2}5$

$g_{2}5≈2.322$

$0−2.322$

SubTerm

Subtract term

$-2.322$

c Recall the Power Property of Logarithms and the Product Property of Logarithms.

$Power:Product: g_{b}m_{n}=ng_{b}mg_{b}mn=g_{b}m+g_{b}n $

Also, as a consequence of the definition of a logarithm, it is known that $g_{b}b=1$ for any positive number $b$ different than $1.$ Using this information, the above properties, and the fact that $g_{2}3≈1.585,$ the given expression can be evaluated.
$g_{2}24$

SplitIntoFactors

Split into factors

$g_{2}(8⋅3)$

$g_{2}(mn)=g_{2}(m)+g_{2}(n)$

$g_{2}8+g_{2}3$

WritePow

Write as a power

$g_{2}2_{3}+g_{2}3$

$g_{2}(a_{m})=m⋅g_{2}(a)$

$3g_{2}2+g_{2}3$

$g_{2}(2)=1$

$3(1)+g_{2}3$

IdPropMult

Identity Property of Multiplication

$3+g_{2}3$

$g_{2}3≈1.585$

$3+1.585$

AddTerms

Add terms

$4.585$

Example

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!
### Hint

### Solution

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 $g_{5}y_{2}25x_{3} $

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x","y"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["2+3\\log_5 x-2\\log_5 y","2-2\\log_5 y+3\\log_5 x","3\\log_5 x-2\\log_5 y+2","3\\log_5 x+2-2\\log_5 y","-2\\log_5 y+2+3\\log_5 x","-2\\log_5 y+3\\log_5 x+2"]}}

b $g_{4}4xy $

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x","y"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["\\dfrac{1}{2} \\log_4 x+\\dfrac{1}{2} \\log_4 y-\\dfrac{1}{2}","\\dfrac{1}{2} \\log_4 x-\\dfrac{1}{2}+\\dfrac{1}{2} \\log_4 y","\\dfrac{1}{2} \\log_4 y+\\dfrac{1}{2} \\log_4 x-\\dfrac{1}{2}","\\dfrac{1}{2} \\log_4 y-\\dfrac{1}{2}+\\dfrac{1}{2} \\log_4 x","-\\dfrac{1}{2}+\\dfrac{1}{2} \\log_4 x+\\dfrac{1}{2} \\log_4 y","-\\dfrac{1}{2}+\\dfrac{1}{2} \\log_4 y+\\dfrac{1}{2} \\log_4 x"]}}

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.

a Consider the Properties of Logarithms.

$Product:Quotient:Power: g_{b}mn=g_{b}m+g_{b}ng_{b}nm =g_{b}m−g_{b}ng_{b}m_{n}=ng_{b}m $

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.
$g_{5}y_{2}25x_{3} $

$g_{5}(ba )=g_{5}(a)−g_{5}(b)$

$g_{5}25x_{3}−g_{5}y_{2}$

$g_{5}(mn)=g_{5}(m)+g_{5}(n)$

$g_{5}25+g_{5}x_{3}−g_{5}y_{2}$

$g_{5}(a_{m})=m⋅g_{5}(a)$

$g_{5}25+3g_{5}x−2g_{5}y$

Calculate logarithm

$2+3g_{5}x−2g_{5}y$

b To use the Power Property of Logarithms, the square root will be written as a rational exponent.

$g_{4}4xy =g_{4}(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.
$g_{4}(4xy )_{21}$

$g_{4}(a_{m})=m⋅g_{4}(a)$

$21 g_{4}4xy $

$g_{4}(ba )=g_{4}(a)−g_{4}(b)$

$21 (g_{4}xy−g_{4}4)$

$g_{4}(mn)=g_{4}(m)+g_{4}(n)$

$21 (g_{4}x+g_{4}y−g_{4}4)$

Distr

Distribute $21 $

$21 g_{4}x+21 g_{4}y−21 g_{4}4$

$g_{4}(4)=1$

$21 g_{4}x+21 g_{4}y−21 (1)$

IdPropMult

Identity Property of Multiplication

$21 g_{4}x+21 g_{4}y−21 $

Example

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.
### Hint

### Solution

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 $g_{2}12+3g_{2}5−g_{2}6$

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["\\log_2 250"]}}

b $5g_{3}2−6g_{3}x+2g_{3}y$

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x","y"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["\\log_3 \\dfrac{32y^2}{x^6}","\\log_3 (\\dfrac{32y^2}{x^6})"]}}

a Use the Product Property of Logarithms, the Power Property of Logarithms, and the Quotient Property of Logarithms.

b Use the Properties of Logarithms.

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: g_{b}mn=g_{b}m+g_{b}ng_{b}nm =g_{b}m−g_{b}ng_{b}m_{n}=ng_{b}m $

These three properties will be used to condense the expression. $g_{2}12+3g_{2}5−g_{2}6$

$m⋅g_{2}(a)=g_{2}(a_{m})$

$g_{2}12+g_{2}5_{3}−g_{2}6$

CalcPow

Calculate power

$g_{2}12+g_{2}125−g_{2}6$

$g_{2}(m)+g_{2}(n)=g_{2}(mn)$

$g_{2}(12⋅125)−g_{2}6$

Multiply

Multiply

$g_{2}1500−g_{2}6$

$g_{2}(m)−g_{2}(n)=g_{2}(nm )$

$g_{2}61500 $

CalcQuot

Calculate quotient

$g_{2}250$

b Similar to Part A, to condense this expression, the same three Properties of Logarithms will be applied.

$5g_{3}2−6g_{3}x+2g_{3}y$

$m⋅g_{3}(a)=g_{3}(a_{m})$

$g_{3}2_{5}−g_{3}x_{6}+g_{3}y_{2}$

CalcPow

Calculate power

$g_{3}32−g_{3}x_{6}+g_{3}y_{2}$

$g_{3}(m)−g_{3}(n)=g_{3}(nm )$

$g_{3}x_{6}32 +g_{3}y_{2}$

$g_{3}(m)+g_{3}(n)=g_{3}(mn)$

$g_{3}(x_{6}32 ⋅y_{2})$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$g_{3}x_{6}32y_{2} $

Closure

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 | $g_{b}a=c⇔a=b_{c}$ |
---|---|

Identity Derived From the Definition | $g_{b}b=1$ |

Identity Derived From the Definition | $g_{b}1=0$ |

Product Property of Logarithms | $g_{b}mn=g_{b}m+g_{b}n$ |

Quotient Property of Logarithms | $g_{b}nm =g_{b}m−g_{b}n$ |

Power Property of Logarithms | $g_{b}m_{n}=ng_{b}m$ |

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.

Loading content