6.4: Algebraic Analysis on the Properties of Logarithms

Learning Objectives

Apply the product rule
Apply the quotient rule
Apply the power rule
Apply the change of base rule
Explain natural logarithms and common logarithms

SInce logarithms are exponents, they have a list of rules that go along with them.

The Identity Property for Logarithms

The definition of logarithm states that $\log_b{a}=x$ is equivalent to $a=b^x$ . Consequently, since $\log_b{x}=\log_b{x}$ then $x=b^{\log_b{x}}$ .

Identity property for Logarithms

For all real numbers $b>0, b\neq1$ ,

$b^{\log_b{x}}=x$

Example 1

Evaluate:

$2^{\log_2{5}}$
$3^{\log_3{7}}$
$7^{\log_7{2}}$

Solution

SInce $b^{\log_b{x}}=x$ ,

$2^{\log_2{5}}=5$
$3^{\log_3{7}}=7$
$7^{\log_7{2}}=2$

Try It 1

Evaluate:

$2^{\log_2{3}}$
$9^{\log_9{7}}$
$2^{\log_2{5}}$

Show Answer

The Product Property for Logarithms

The product Property for logarithms mimics the product Property for exponents. SInce logarithms are exponents the exponential property $a^m\cdot a^n=a^{m+n}$ gets translated into logarithmic form. The multiplication of terms inside the argument of a logarithm is equal to the addition of logarithms of each term.

The Product Property for logarithms

For any positive real numbers $b,\;M,\;N$ , and $b\neq1$ ,

$\log_b{MN} = \log_b{M}+\log_b{N}$

Example 2

Expand the logarithm $\log_3{3x}$ .

Solution

The argument consists of the product of 3 and $x$ , so applying the product property:

$\log_3{3x}=\log_3{3}+\log_3{x}$

We can then evaluate $\log_3{3}=1$ , to get:

$\log_3{3}+\log_3{x}=1+\log_3x$

Example 3

Expand the logarithm $\log_5{[5s(t-3)]}$ .

Solution

The argument consists of the product of $5\cdot s\cdot (t-3)$ , so we can expand the product property:

$\begin{aligned}\log_5{[5s(t-3)]} &= \log_5{5}+\log_5{s} +\log_5{(t-3)}\\&=1+\log_5{s}+\log_5{(t-3)}\end{aligned}$

Try It 2

Expand the logarithm $\log_4{16x}$ .

Show Answer

$2+\log_4x$

Try It 3

Expand the logarithm $\log_2{[8(s+2)(t-5)]}$ .

Show Answer

$3+\log_2(s+2)+\log_2(t-5)$

The Quotient Property for Logarithms

The quotient property for logarithms mimics the quotient property for exponents. SInce logarithms are exponents the exponential property $\dfrac{a^m}{a^n}=a^{m-n}$ gets translated into logarithmic form. The division of terms inside the argument of a logarithm is equal to the subtraction of logarithms of each term.

The quotient Property for logarithms

For any positive real numbers $b,\;M,\;N$ , and $b\neq1$ ,

$\log_b{\dfrac{M}{N}} = \log_b{M}-\log_b{N}$

Example 4

Expand the logarithm $\log_{10}{\frac{x}{10}}$ .

Solution

The argument of the log consists of a quotient of the two terms $x$ and 10, so we apply the quotient property:

$\begin{aligned}\log_{10}{\frac{x}{10}} &= \log_{10}{x}-\log_{10}{10} \\&=\log_{10}{x}-1\end{aligned}$

Try It 4

Expand the logarithm $\log_{4}{\frac{x}{4}}$ .

Show Answer

$\log_4x-1$

The Power Property for Logarithms

The power Property comes from the product rule of exponents. $\log_b{M^r}=\log_b{(M\cdot M\cdot M...M\cdot M)}=\log_bM+\log_bM+\log_bM+...+\log_bM=r\log_bM$ . If the argument of a logarithm is a single term with an exponent, we may pull the exponent in front of the logarithm.

The Power Property for Logarithms

For any positive real numbers $b,\;M,\;N$ , and $b\neq1$ ,

$\log_b{M^r}=r\times\log_b{M}$

Example 5

Expand the logarithm $\log_{2}{x^4}$ .

Solution

The argument of the logarithm is $x^4$ , a single term raised to a power, so we can apply the power property of logarithms:

$\log_{2}{x^4}=4\log_2{x}$

Try It 5

Expand the logarithm $\log_{7}{x^{-2}}$ .

Show Answer

$-2\log_7x$

Expanding a Logarithm

The properties we have learned so far to expand a logarithm can be used in conjunction with one another.

Example 6

Expand the logarithm $\log_{2}{\frac{2y}{5}}$ .

Solution

$\begin{aligned}\log_{2}{\frac{2y}{5}} &=\log_2{(2y)}-\log_2{5}&&\text{Quotient Property}\\&=\log_2{2}+\log_2{y}-\log_2{5}&&\text{Product Property}\\&=1+\log_2{y}-\log_2{5}&&\text{IdentityProperty}\end{aligned}$

Example 7

Expand the logarithm $\log_{4}{\frac{16y^3}{5}}$ .

Solution

$\begin{aligned}\log_{4}{\frac{16y^3}{5}} &=\log_4{(16y^3)}-\log_4{5}&&\text{Quotient Property}\\&=\log_4{16}+\log_4{y^3}-\log_4{5}&&\text{Product Property}\\&=\log_4{4^2}+3\log_4{y}-\log_4{5}&&\text{Power Property}\\&=2+3\log_4{y}-\log_4{5}&&\text{Identity Property}\end{aligned}$

Try It 6

Expand the logarithm $\log_{10}{100x^3y^5}$ .

Show Answer

$2+3\log_{10}x+5\log_{10}y$

Try It 7

Expand the logarithm $\log_{2}{\dfrac{32x}{y^5}}$ .

Show Answer

$5+\log_2x-5\log_2y$

Combining Logarithms

The properties for exponents we applied to expand a logarithm can be also be used (in reverse) to combine logarithms into a single logarithm. The assumption is that these logarithms have the same base.

Example 8

Combine $\log_4{2}+\log_4{10}-log_4{5}$ into a single logarithm.

Solution

$\begin{aligned}\log_4{2}+\log_4{10}-\log_4{5}&=\log_4{(2\times10)}-\log_4{5}&&\text{Product Property}\\&=\log_4{\frac{20}{5}}&&\text{Quotient Property}\\&=\log_4{4}&&\text{Simplify}\\&=1\end{aligned}$

Example 9

Combine $2\log_2{(x-2)}-\log_2{(x^2-x-2)}$ into a single logarithm.

Solution

$\begin{aligned}2\log_2{(x-2)}-\log_2{(x^2-x-2)}&=\log_2{(x-2)^2}-\log_2{(x^2-x-2)}&&\text{Power Property}\\&=\log_2{\frac{(x-2)^2}{x^2-x-2}}&&\text{Quotient Property}\\&=\log_2{\frac{(x-2)^2}{(x-2)(x+1)}}&&\text{Factor}\\&=\log_2{\frac{(x-2)}{(x+1)}}&&\text{Simplify}\end{aligned}$

Try It 8

Combine $\log_2{3}+4\log_2{x}+5\log_2{y}$ into a single logarithm.

Show Answer

$\log_2{3x^4y^5}$

Try It 9

Combine $4\log_2{x}-\log_2{7}-5\log_2{y}$ into a single logarithm.

Show Answer

$\log_2{\dfrac{x^4}{7y^5}}$

Try It 10

Combine $2\log_2{(x-1)}-\log_2{(x^2-1)}$ into a single logarithm.

Show Answer

$\log_2{\dfrac{x-1}{x+1}}$

Common and Natural Logarithms

If the base of a logarithm is 10, the logarithm is call a common logarithm. This is because the common number system we use in our daily life is base-10. If the base of a logarithm is the number $e$ (also known as Euler’s number), the logarithm is call a natural logarithm. This is because the number $e$ is an irrational number equal to 2.718281828459…; the ellipses … mean that the decimal never ends and never repeats. The number $e$ was first discovered in 1683 by the Swiss mathematician Jacob Bernoulli but was first evaluated by Leonhard Euler in 1737. Consequently, it is often referred to as Euler’s number. The natural logarithm is one of the most useful functions in mathematics, with applications throughout the physical and biological sciences.

Calculator buttons

Most scientific calculators have both $\log_{10}$ , which is written as $\log$ (the base is so common it is omitted) and $\log_e$ which is written $\ln$ ( $\ln$ refers to logarithm naturale).

Notice that behind the log button is $10^x$ , and behind the ln button is $e^x$ . These are the inverse functions.

Example 10

Use a calculator to evaluate the logarithms exactly or to 5 decimal places:

$\log4$
$\log100$
$\ln4$
$\ln{e}$

Solution

$\log4=0.60206$
$\log100=2$
$\ln{4}=1.38629$
$\ln{e}=1$

All of the properties of logarithms apply to common and natural logarithms.

Properties of Logarithms

Property	Base 10	Base $e$
Identity	$\log{10}=1$	$\ln{e}=1$
Product	$\log{MN}=\log{M}+\log{N}$	$\ln{MN}=\ln{M}+\ln{N}$
Quotient	$\log{\dfrac{M}{N}}=\log{M}-\log{N}$	$\ln{\dfrac{M}{N}}=\ln{M}-\ln{N}$
Power	$\log{M^r}=r\log{M}$	$\ln{M^r}=r\ln{M}$

Example 11

Evaluate $\ln{e^2}$ .

Solution

$\begin{aligned}\ln{e^2}&=2\ln{e}\\&=2\times1\\&=2\end{aligned}$

Try It 11

Evaluate $\log{10,000}$ .

Show Answer

$4$

Example 12

Simplify to a single logarithm: $\log{(x^2+4x+3)}-2\log{(x+1)}$

Solution

$\begin{aligned}\log{(x^2+4x+3)}-2\log{(x+1)}&=\log{(x+3)(x+1)}-\log{(x+1)^2}&&\text{Factor. Power property.}\\&=\log{\dfrac{(x+3)(x+1)}{(x+1)^2}}&&\text{Quotient property}\\&=\log{\dfrac{x+3}{x+1}}&&\text{Simplify}\end{aligned}$

Try It 12

Simplify to a single logarithm: $4\ln{x}-2\ln{(x+1)}+\ln{(x^2-1)}$

Show Answer

$\ln{\dfrac{x^4(x-1)}{(x+1)}}$

The Change of Base Rule

Calculators are a great way to evaluate logarithms with base 10 or $e$ . But what if we want to evaluate a logarithm with a different base? The good news is that we can change the base of any logarithm to any base we wish to use, most importantly base 10 and $e$ .

The Change of Base Formula

For any positive real number $m$ with $m\neq1$ ,

$\log_b{a}=\dfrac{\log_m{a}}{\log_m{b}}$

SInce we usually have access to a calculator with logarithms of base 10 and $e$ , it is important to be able to change the base of any logarithm to 10 or $e$ . The change of base formula can be used:

$\log_ba=\dfrac{\log{a}}{\log{b}}$ and $\log_ba=\dfrac{\ln{a}}{\ln{b}}$

Example 13

Evaluate $\log_2{5}$ to 5 decimal places.

Solution

Converting to common logs:

$\begin{aligned}\log_2{5}&=\dfrac{\log{5}}{\log{2}}\\&=2.32193\end{aligned}$

$\begin{aligned}\log_2{5}&=\dfrac{\ln{5}}{\ln{2}}\\&=2.32193\end{aligned}$

Notice that when using a calculator we never write the answer to the numerator and denominator of a fraction, then divide those two rounded numbers. This is because rounding each number than dividing brings in rounding error. Instead we complete the full calculation in the calculator, then round at the very end.

Try It 13

Evaluate $\log_9{32}$ to 5 decimal places.

Show Answer

$1.57732$

Composition with the Inverse Function (Application of the Properties of Logarithms)

In chapter 3.3.2, we learned that $f(x)$ and $g(x)$ are inverses of each other if $f\left(g\left(x\right)\right)=x$ and $g\left(f\left(x\right)\right)=x$ . We may use this property to determine if the inverse function we found is correct or not. For example, suppose someone found that the inverse function of $f(x)=\log_2{(x+3)}-5$ is $f^{-1}(x)=2^{x+5}-3$ . We may justify this answer by checking if $f(f^{-1}(x))=x$ and $f^{-1}(f(x))=x$ . If either one is false, then the inverse function is incorrect.

$\begin{aligned}f(f^{-1}(x))&=\log_2{((2^{x+5}-3)+3)}-5\\&=\log_2{2^{x+5}}-5\\&=(x+5)\log_2{2}-5\\&=(x+5)\times1-5\\&=x+5-5\\&=x\end{aligned}$

$\begin{aligned}f^{-1}(f(x))&=2^{(\log_2{(x+3)}-5)+5}-3\\&=2^{(\log_2{(x+3)}}-3\\&=(x+3)-3\\&=x\end{aligned}$

Therefore, the inverse function $f^{-1}(x)=2^{x+5}-3$ is correct because $f(f^{-1}(x))=x$ and $f^{-1}(f(x))=x$ .

Chapter 6: Logarithmic Functions