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College Algebra/Logarithmic Functions/Properties of Logarithms

Logarithmic Functions

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Properties of Logarithms

Properties of logarithms can be used to simplify logarithmic expressions.
Logarithms have several important properties that can be used to combine them or write them in different forms.

Logarithm Properties

Name Property
Product rule of logarithms
log⁡b(xy)=log⁡bx+log⁡by\log_b{(xy)}=\log_b{x}+\log_b{y}
Quotient rule of logarithms
log⁡b(xy)=log⁡bx−log⁡by\log_b{\left (\frac{x}{y} \right )}=\log_b{x}-\log_b{y}
Power rule of logarithms
log⁡bxp=p⋅log⁡bx\log_b{x^p}=p\cdot \log_b{x}
Change of base rule
log⁡bx=log⁡axlog⁡ab\log_b{x}=\frac{\log_a{x}}{\log_a{b}}

There are also two important forms to remember. When the value of the argument is 1 and the base is a positive value other than 1, then the logarithm is equal to zero:
log⁡b1=0\log_b{1}=0
When the value of the base is the same as the value of the argument and the vase is a positive value other than 1, then the logarithm is equal to 1:
log⁡bb=1\log_b{b}=1

Logarithm of a Product

To determine the logarithm of a product, add the logarithms of the factors.
The product rule of logarithms states that the logarithm of a product is equal to the sum of the logarithms of the factors. In other words, where bb, xx, and yy are positive real numbers and b≠1b\neq 1, the product rule of logarithms can be written as:
log⁡b(xy)=log⁡bx+log⁡by\log_b{(xy)}=\log_b{x}+\log_b{y}
Like other logarithmic properties, the product rule of logarithms can be used to evaluate logarithmic expressions or to solve complex logarithmic equations.
Step-By-Step Example
Adding Logarithms to Simplify an Expression
Rewrite the given expressionas a single logarithm, and then find the value of the given expression:
log⁡612+log⁡618\log_6{12}+\log_6{18}
Step 1
The logarithms have the same base, so use the product rule.
log⁡bx+log⁡by=log⁡b(xy)\log_b{x}+\log_b{y}=\log_b{(xy)}
The given expression can be rewritten as:
log⁡612+log⁡618log⁡6(12⋅18)\begin{gathered}\log_6{12}+\log_6{18}\\\log_6{(12\cdot 18)}\end{gathered}
Step 2
Multiply.
log⁡6(12⋅18)log⁡6216\begin{gathered}\log_6{(12\cdot 18)}\\\log_6{216}\end{gathered}
Step 3

Use the relationship between exponents and logarithms to evaluate log⁡6216\log_6{216}.

Write 216 as a power of 6.
216=6⋅6⋅6=63\begin{aligned}216&=6\cdot 6\cdot 6\\&=6^{3}\end{aligned}
The exponential equation 63=2166^3=216 can be rewritten as the equivalent logarithmic equation:
log⁡6216=3\log_6{216}=3
Solution
The value of the given logarithmic expression is 3.
Step-By-Step Example
Evaluating the Logarithm of a Product
Evaluate the given logarithm:
log⁡5(25⋅625)\log_5{(25 \cdot 625)}
Step 1
Use the product rule to rewrite the logarithm of a product as the sum of the logarithms of the factors.
log⁡b(xy)=log⁡bx+log⁡bylog⁡5(25⋅625)=log⁡525+log⁡5625\begin{aligned}\log_b{(xy)}&=\log_b{x}+\log_b{y}\\\log_5{(25 \cdot 625)}&=\log_5{25}+\log_5{625}\end{aligned}
Step 2

Evaluate the first logarithm. Use aa as the exponent.

The exponential form of log⁡525\log_5{25} is:
5a=255^a=25
Since 52=255^2=25, log⁡525\log_{5}25 is 2:
log⁡525=2\log_5{25}=2
Step 3

Evaluate the second logarithm. Use bb as the exponent.

The exponential form of log⁡5625\log_5{625} is:
5b=6255^b=625
Since 54=6255^4=625, log⁡5625\log_5{625} is 4:
log⁡5625=4\log_5{625}=4
Step 4
Calculate the sum of the logarithms from Step 2 and Step 3.
log⁡525+log⁡5625=2+4=6\begin{aligned}\log_5{25}+\log_5{625}&=2+4\\&=6\end{aligned}
Solution
The value of the given logarithm is 6:
log⁡5(25⋅625)=6\log_5{(25 \cdot 625)}=6

Logarithm of a Quotient

To calculate the logarithm of a quotient, subtract the logarithm of the divisor from the logarithm of the dividend.
The quotient rule of logarithms states that the logarithm of a quotient is equal to the difference of the logarithm of the dividend and the logarithm of the divisor. To calculate the logarithm of a quotient, use the quotient rule of logarithms, where bb, xx, and yy are positive and b≠1b\neq 1, by subtracting the logarithm of the divisor from the logarithm of the dividend:
log⁡b(xy)=log⁡bx−log⁡by\log_b{\left (\frac{x}{y} \right )}=\log_b{x}-\log_b{y}
Step-By-Step Example
Evaluating the Logarithm of a Quotient
Evaluate the given logarithm:
log⁡2(864)\log_2{\left (\frac{8}{64} \right )}
Step 1
Use the quotient rule to rewrite the logarithm of a quotient as a difference of logarithms.
log⁡b(xy)=log⁡bx−log⁡bylog⁡2(864)=log⁡28−log⁡264\begin{aligned}\log_b\left (\frac{x}{y} \right )&=\log_b{x}-\log_b{y}\\ {\log_2{{\left (\frac{8}{64} \right )}}}&=\log_2{8}-\log_2{64}\end{aligned}
Step 2

Evaluate the first logarithm. Use aa as the exponent.

The exponential form of log⁡28\log_{2}8 is:
2a=82^{a}=8
Since 23=82^{3}=8, log⁡28=3\log_{2}8=3 is 3:
log⁡28=3\log_2{8}=3
Step 3

Evaluate the second logarithm. Use bb as the exponent.

The exponential form of log⁡264\log_{2}64 is:
2b=642^{b}=64
Since 26=642^6=64, log⁡264\log_{2}64 is 6:
log⁡264=6\log_2{64}=6
Step 4
Calculate the difference.
log⁡28−log⁡264=3−6=−3\begin{aligned}\log_2{8}-\log_2{64}&=3-6\\&=-3\end{aligned}
Solution
The value of the given logarithm is –3.
log⁡2(864)=−3\log_2{\left (\frac{8}{64} \right )}=-3
Step-By-Step Example
Evaluating the Logarithm of a Quotient That Contains a Sum
Simplify:
log⁡7(7x3+5)\log_7{\left (\frac{7}{x^{^{3}}+5} \right )}
Step 1
Use the quotient rule to rewrite the logarithm of a quotient as a difference of logarithms.
log⁡b(xy)=log⁡bx−log⁡bylog⁡7(7x3+5)=log⁡77−log⁡7(x3+5)\begin{aligned}\log_b{\left (\frac{x}{y} \right )}&=\log_b{x}-\log_b{y}\\\log_7{\left (\frac{7}{x^{^{3}}+5} \right )}&=\log_7{7}-\log_7{(x^3+5)}\end{aligned}
Step 2

Examine the first term, log⁡77\log_{7}7.

The value of the base is the same as the value of the argument. The base is also a positive value other than 1. So, the logarithm is equal to zero:

Since log⁡bb=1\log_b{b}=1 for any positive base that is not equal to 1, the first term is equal to 1.
log⁡bb=1log⁡77=1\begin{aligned}\log_b{b}&=1\\\log_7{7}&=1\end{aligned}
Step 3
Examine the second term:
log⁡7(x3+5)\log_7{(x^3+5)}
There is no rule that allows the second term to be further simplified because it is the logarithm of a sum.
Solution
Put the examined terms together:
log⁡7(7x3+5)=1−log⁡7(x3+5)\log_7{\left (\frac{7}{x^{^{3}}+5} \right )}=1-\log_7{(x^3+5)}

Logarithm of a Power

To determine the logarithm of a power, multiply the exponent by the logarithm of the base.
The power rule of logarithms is equal to the product of the exponent and the logarithm of the base. In other words, where pp is a real number, bb and xx are positive, and b≠1b\neq 1, the logarithm of a power can be written as:
log⁡bxp=p⋅log⁡bx\log_b{x^p}=p\cdot \log_b{x}
To identify the logarithm of a power, multiply the exponent of the power by the logarithm of the base. For example:
log⁡3115=5⋅log⁡311\log_3{11^5}=5\cdot \log_3{11}
The power rule and other properties of logarithms can be applied to write logarithmic expressions in expanded form.
Step-By-Step Example
Using Properties to Expand Expressions
Write the given expression in expanded form:
log⁡3(4x53y7)\log_3{\left ( \frac{4x^{5}}{3y^{7}} \right)}
Step 1
Use the quotient rule of logarithms to rewrite the logarithm of a quotient as a difference of logarithms.
log⁡b(xy)=log⁡bx−log⁡bylog⁡3(4x53y7)=log⁡3(4x5)−log⁡3(3y7)\begin{aligned}\log_b{\left (\frac{x}{y} \right )}&=\log_b{x}-\log_b{y}\\\log_3{\left ( \frac{4x^{5}}{3y^{7}} \right )}&=\log_3{(4x^5)}-\log_3{(3y^7)}\end{aligned}
Step 2
The product rule of logarithms is:
log⁡b(xy)=log⁡bx+log⁡by\log_b{(xy)}=\log_b{x}+\log_b{y}
Rewrite each logarithm of a product as a sum of logarithms and simplify.
log⁡3(4x5)⏞First logarithm−log⁡3(3y7)⏞Second logarithmlog⁡34+log⁡3(x5)⏞First logarithm−[log⁡33+log⁡3(y7)]⏞Second logarithmlog⁡34+log⁡3(x5)⏞First logarithm−log⁡33−log⁡3(y7)⏞Second logarithm\begin{gathered}&\overbrace{\log_{3}(4x^5)}^{\text{First logarithm}}-\overbrace{\log_3{(3y^7)}}^{\text{Second logarithm}}\\&\overbrace{\log_{3}4+\log_3{(x^5)}}^{\text{First logarithm}}-\overbrace{[\log_{3}3+\log_{3}(y^7)]}^{\text{Second logarithm}}\\&\overbrace{\log_{3}4+\log_{3}(x^5)}^{\text{First logarithm}}-\overbrace{\log_{3}3-\log_{3}(y^{7})}^{\text{Second logarithm}}\end{gathered}
Step 3
The power rule of logarithms is:
log⁡bxp=p⋅log⁡bx\log_b{x^p}=p\cdot \log_b{x}
Use the power rule of logarithms to rewrite the terms with exponents:
log⁡34+log⁡3(x5)⏞First term−log⁡33−log⁡3(y7)⏞Second termlog⁡34+5⋅log⁡3x⏞First term−log⁡33−7⋅log⁡3y⏞Second term\begin{gathered}\log_{3}4+\overbrace{\log_{3}(x^5)}^\text{First term}-\log_{3}3-\overbrace{\log_{3}(y^7)}^{\text{Second term}}\\\log_{3}4+\overbrace {5\cdot\log_{3}x}^\text{First term}-\log_{3}3-\overbrace{7\cdot \log_{3}y}^{\text{Second term}}\end{gathered}
Step 4
Since any positive number raised to the power of 1 is equal to that number, it is also true that:
log⁡bb=1log⁡33=1\begin{aligned}\log_b{b}&=1\\\log_3{3}&=1\end{aligned}
Substitute 1 for log⁡33\log_3{3}.
log⁡34+5⋅log⁡3x−log⁡33−7⋅log⁡3ylog⁡34+5⋅log⁡3x−1−7⋅log⁡3y\begin{gathered}\log_3{4}+5\cdot \log_3{x}-{\color{#c42126}{\log_3{3}}}-7\cdot \log_3{y}\\\log_3{4}+5\cdot \log_3{x}-{\color{#c42126}{1}}-7\cdot \log_3{y}\end{gathered}
Solution
The expanded form of the given expression is:
log⁡34+5⋅log⁡3x−1−7⋅log⁡3y\log_3{4}+5\cdot\log_3{x}-1-7\cdot \log_3{y}

Change of Base

A logarithm can be changed from one base to another by using the formula log⁡bx=log⁡axlog⁡ab\log_b{x}=\frac{\log_a{x}}{\log_a{b}}.
It can sometimes be convenient to change a logarithm from one base to another. The change of base rule states that a logarithm of a number in base bb can be changed to base aa by dividing the logarithm of the number in base aa by the logarithm of bb in base aa. The change of base rule, where aa, bb, and xx are positive, a≠1a\neq 1, and b≠1b \neq 1, can be written as:
log⁡bx=log⁡axlog⁡ab\log_b{x}=\frac{\log_a{x}}{\log_a{b}}
It shows how to change a logarithm in base bb to a different base aa. This rule is commonly used to calculate the values of logarithms with bases other than base 10 or ee when using technology, in particular, calculators with only the common log and the natural log. To use the change of base rule with a calculator, select either log or ln and calculate the quotient of the log of the argument and the log of the original base. For example:
log⁡512=log⁡12log⁡5=ln⁡12ln⁡5≈1.544\log_5{12}=\frac{\log{12}}{\log{5}}=\frac{\ln{12}}{\ln{5}}\approx1.544
Step-By-Step Example
Deriving the Change of Base Rule
Derive the change of base rule:
log⁡bx=log⁡axlog⁡ab\log_b{x}=\frac{\log_a{x}}{\log_a{b}}
Step 1

The change of base rule shows how to write log⁡bx\log_{b}x in terms of a logarithm with another base, aa.

First, let yy represent the logarithm being derived, or log⁡bx\log_{b}x:
y=log⁡bxy=\log_b{x}
Step 2
Next, rewrite the logarithm from Step 1 in exponential form.
y=log⁡bxx=by\begin{aligned}y&=\log_b{x}\\x&=b^y\end{aligned}
Step 3
Take log⁡a\log_a of both sides of the equation.
x=bylog⁡ax=log⁡a(by)\begin{aligned}x&=b^y\\ \log_a{x}&=\log_a{(b^y)}\end{aligned}
Step 4
Simplify the logarithm using the power rule of logarithms by multiplying the exponent by the logarithm of the base:
log⁡ax=log⁡a(by)log⁡ax=y⋅log⁡ab\begin{aligned}\log_a{x}&=\log_a{(b^y)}\\\log_a{x}&=y\cdot \log_a{b}\end{aligned}
Step 5
Solve for yy by dividing both sides by log⁡ab\log_a{b}.
log⁡ax=y⋅log⁡ablog⁡axlog⁡ab=y\begin{aligned}\log_a{x}&=y \cdot \log_a{b}\\\frac{\log_a{x}}{\log_a{b}}&=y\end{aligned}
Solution
Remember that y=log⁡bxy=\log_b{x} from Step 1. Substitute yy for log⁡bx\log_b{x}:
y=log⁡axlog⁡ablog⁡bx=log⁡axlog⁡ab\begin{aligned}y&=\frac{\log_a{x}}{\log_a{b}}\\\log_b{x}&=\frac{\log_a{x}}{\log_a{b}}\end{aligned}
Step-By-Step Example
Changing the Base of a Logarithm
Evaluate log⁡432\log_4{32}.
Step 1

Identify a different base.

The argument 32 is not a whole-number power of the base 4, but notice that both the argument and the base are powers of 2.

Change the logarithm to a base of 2.

Step 2
The change of base rule is:
log⁡bx=log⁡axlog⁡ab\log_b{x}=\frac{\log_a{x}}{\log_a{b}}
Apply the change of base rule to the logarithm.
log⁡432=log⁡232log⁡24\log_4{32}=\frac{\log_2{32}}{\log_2{4}}
Step 3

Evaluate the logarithm in the numerator. Use aa as the exponent.

The exponential form of log⁡232\log_2{32} is:
2a=322^{a}=32
Since 25=322^5=32, the logarithm in the numerator is 5:
log⁡232=5\log_2{32}=5
Step 4

Evaluate the logarithm in the denominator. Use bb as the exponent.

The exponential form of log⁡24\log_2{4} is:
2b=42^{b}=4
Since 22=42^{2}=4, the logarithm in the denominator is 2:
log⁡24=2\log_2{4}=2
Step 5
Substitute the simplified logarithms back into the logarithm from Step 1:
log⁡232log⁡24=52\frac{\log_2{32}}{\log_2{4}}=\frac{5}{2}
Solution
The value of the given logarithm is 52\frac{5}{2}:
log⁡432=52\log_4{32}=\frac{5}{2}
<Relating Exponents to Logarithms>Graphing Logarithmic Functions