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{{ printedBook.courseTrack.name }} {{ printedBook.name }} The Properties of Logarithms allow expressions with logarithms to be rewritten.

The logarithm of a product can be written as the sum of the logarithms of each factor in the product. For example, $g(7⋅4)$ can be expressed using this rule.
$g(7⋅4)=g(7)+g(4)$
This rule can be explained using the identity $x=b_{log_{b}(x)}.$
This rule is valid for positive values of $b,$ $m,$ and $n$ and $b =1.$

$g_{b}(mn)$

$g_{b}(m⋅n)$

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

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

MultPow$a_{m}⋅a_{n}=a_{m+n}$

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

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

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

The logarithm of a quotient can be written as the difference between the logarithm of the numerator and the logarithm of the denominator. For example, $g(47 )$ can be expressed using this rule.
$g(47 )=g(7)−g(4)$
This rule can be explained using the identity $x=b_{log_{b}(x)}.$
This rule is valid for positive values of $b,$ $m,$ and $n$ and $b =1.$

$g_{b}(nm )$

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

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

DivPow$a_{n}a_{m} =a_{m−n}$

$g_{b}(b_{log_{b}(m)−log_{b}(n)})$

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

$g_{b}(m)−g_{b}(n)$

The logarithm of a power can be written as the product of the exponent and the logarithm of the base. For example, $g(7_{4})$ can be expressed using this rule.
$g(7_{4})=4⋅g(7)$
This rule can be explained using the identity $x=b_{log_{b}(x)}.$
This rule is valid for positive values of $b,$ $m,$ and $n$ and $b =1.$

$g_{b}(m_{n})$

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

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

PowPow$(a_{m})_{n}=a_{m⋅n}$

$g_{b}(b_{log_{b}(m)⋅n})$

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

$g_{b}(m)⋅n$

CommutativePropMultCommutative Property of Multiplication

$ng_{b}(m)$

Simplify the expression $g_{6}(8)−g_{6}(3)+g_{6}(9)−g_{6}(4)$ using the properties of logarithms.

Show Solution

The Quotient Property of Logarithms states that the following is true for logarithms.
$g_{b}(nm )=g_{b}(m)−g_{b}(n)$
In our expression the first two terms and the last two terms both are a difference of two logarithms with the same base. We can use the Quotient Property of Logarithms to rewrite each pair into a logarithm of a quotient. $g_{6}(8)−g_{6}(3)+g_{6}(9)−g_{6}(4)⇕g_{6}(38 )+g_{6}(49 )$
This expression is a sum of two logarithms with the same base. We can write them into a logarithm of a product using the Product Property of Logarithms.
$g_{6}(38 )+g_{6}(49 )⇔g_{6}(38 ⋅49 )$
Let's continue simplifying this expression.
Thus, the expression $g_{6}(8)−g_{6}(3)+g_{6}(9)−g_{6}(4)$ equals $1.$

$g_{6}(38 ⋅49 )$

MultFracMultiply fractions

$g_{6}(3⋅48⋅9 )$

SimpQuotSimplify quotient

$g_{6}(6)$

$g_{6}(6)=1$

$1$

The Change of Base Formula allows the logarithm of an arbitrary base to be rewritten as the quotient of two logarithms with another base. $g_{c}(a)=g_{b}(c)g_{b}(a) $ With many calculators it is only possible to evaluate the common and the natural logarithm. The Change of Base Formula can then be used to evaluate logarithms of other bases.

$g_{c}(a)=g(c)g(a) andg_{c}(a)=ln(c)ln(a) $
An exponential equation can be solved by applying logarithms and using the Power Property of Logarithms. Consider the following equation.
$8_{x}=3 $
### 1

### 2

By using the Power Property of Logarithms, powers can be rewritten into a product. $g(8_{x})=g(3)⇕x⋅g(8)=g(3) $

### 3

After the power has been rewritten into a product, the unknown variable can be isolated using inverse operations. Here, $x$ gets isolated on the left-hand side when both sides of the equation are divided by $g(8).$ $x⋅g(8)=g(3)⇕x=g(8)g(3) $ By using a calculator, an approximate value of $x$ can be calculated. Here, $x≈0.53.$

Apply the logarithm

First, the equation is rewritten by applying a logarithm on both sides. $8_{x}=3⇔g(8_{x})=g(3)$

Rewrite using the Power Property of Logarithms

Solve the resulting equation

Solve the equation $4_{x}=25$ using the common logarithm. State the answer with three significant digits.

Show Solution

The given equation can be solved by applying the $g_{4}$ on both sides. Since many calculators are limited to only the common and the natural logarithms, it is often not possible to use $g_{4}.$ We can solve it anyway. Let's begin by applying the common logarithm on both sides.
$4_{x}=25⇔g(4_{x})=g(25)$
The Power Property of Logarithms gives us the relationship $g_{b}(m_{n})=n⋅g_{b}(m).$ With this we can rewrite the left-hand side of the equation.
$g(4_{x})=g(25)⇔x⋅g(4)=g(25)$
We can now solve the equation for $x$ by dividing both sides of the equation with $g(4).$
The solution to the equation is $x≈2.32.$

$x⋅g(4)=g(25)$

DivEqn$LHS/g(4)=RHS/g(4)$

$x=g(4)g(25) $

UseCalcUse a calculator

$x=2.32192…$

RoundDecRound to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$

$x≈2.32$

A logarithmic function $g(x)=g_{b}(x),$ is by definition the inverse of an exponential function $f(x)=b_{x}.$ This means that their function composition results in the identity function.

- $g(f(x))=g_{b}(b_{x})=x$
- $f(g(x))=b_{log_{b}(x)}=x$

The fact that a logarithm and a power with the same base undo

each other is what is known as the **inverse properties of logarithms**.

$g_{b}(b_{x})=xandb_{log_{b}(x)}=x $

They also hold true for the common logarithm and the natural logarithm.

$g(10_{x})=xandln(e_{x})=xand 10_{log(x)}=xe_{ln(x)}=x $

These properties together with other properties of logarithms permit to simplify logarithmic expressions and to solve equations involving logarithms and powers. Some particular examples are shown below.

$ln(e_{4x})4xx =20=20=5 3_{log_{3}(5x)}5xx =10=10=2 $

Solve the equation $g_{5}(125)e_{ln(3_{4})}⋅10_{log(x)} =g(100)$ using the inverse properties of logarithms.

Show Solution

The inverse properties of logarithms tell us that a logarithm and a power with the same base undo each other. $g_{b}(b_{x})=xandb_{log_{b}(x)}=x$
On the left-hand side, we can identify $ln$ as being the logarithm with base $e$ and $g$ as the logarithm with base $10.$ Thus, we can simplify the factors $e_{ln(3_{4})}$ and $10_{log(x)}$ using the inverse properties of logarithms.
$g_{5}(125)e_{ln(3_{4})}⋅10_{log(x)} ⇕g_{5}(125)3_{4}⋅x =g(100)=g(100) $
To simplify the remaining two logarithms, we first need to rewrite their arguments. $g_{5}(125)3_{4}⋅x =g(100)⇔g_{5}(5_{3})3_{4}⋅x =g(10_{2})$
Again, we can use the the inverse properties of logarithms to simplify the equation.
$g_{5}(5_{3})3_{4}⋅x =g(10_{2})⇔33_{4}⋅x =2$
We now need to isolate $x$ on the left-hand side to find the solution.
We have solved the equation, and its solution is $272 .$

$33_{4}⋅x =2$

SimpQuotSimplify quotient

$3_{3}⋅x=2$

DivEqn$LHS/3_{3}=RHS/3_{3}$

$x=3_{3}2 $

CalcPowCalculate power

$x=272 $

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