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# Using Properties of Logarithms

Rule

## Properties of Logarithms

The Properties of Logarithms allow expressions with logarithms to be rewritten.

Rule

## Product Property of Logarithms

### Rule

$\log_b(mn)=\log_b(m)+\log_b(n)$
The logarithm of a product can be written as the sum of the logarithms of each factor in the product. For example, $\log(7\cdot4)$ can be expressed using this rule. $\log(7\cdot4)=\log(7)+\log(4)$ This rule can be explained using the identity $x=b^{\log_b(x)}.$
$\log_b(mn)$
$\log_b(m\cdot n)$
$m={b}^{\log_{b}(m)}$
$\log_b\left(b^{\log_b(m)}\cdot b^{\log_b(n)}\right)$
$\log_b\left(b^{\log_b(m)+\log_b(n)}\right)$
$\log_{b}\left({b}^m\right)=m$
$\log_b(m)+\log_b(n)$
This rule is valid for positive values of $b,$ $m,$ and $n$ and $b\neq 1.$
Rule

## Quotient Property of Logarithms

### Rule

$\log_b\left(\dfrac{m}{n}\right)=\log_b(m)-\log_b(n)$
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, $\log\left(\frac{7}{4} \right)$ can be expressed using this rule. $\log\left(\dfrac{7}{4}\right)=\log(7)-\log(4)$ This rule can be explained using the identity $x=b^{\log_b(x)}.$
$\log_b\left(\dfrac{m}{n} \right)$
$m={b}^{\log_{b}(m)}$
$\log_b\left(\dfrac{b^{\log_b(m)}}{b^{\log_b(n)}}\right)$
$\log_b\left(b^{\log_b(m)-\log_b(n)}\right)$
$\log_{b}\left({b}^m\right)=m$
$\log_b(m)-\log_b(n)$
This rule is valid for positive values of $b,$ $m,$ and $n$ and $b\neq 1.$
Rule

## Power Property of Logarithms

### Rule

$\log_b\left(m^n\right)=n\log_b(m)$
The logarithm of a power can be written as the product of the exponent and the logarithm of the base. For example, $\log \left(7^4 \right)$ can be expressed using this rule. $\log \left(7^4 \right)=4\cdot\log(7)$ This rule can be explained using the identity $x=b^{\log_b(x)}.$
$\log_b(m^n)$
$m={b}^{\log_{b}(m)}$
$\log_b\left (\left(b^{\log_b(m)} \right)^n\right)$
$\log_b\left (b^{\log_b(m)\cdot n}\right)$
$\log_{b}\left({b}^m\right)=m$
$\log_b(m)\cdot n$
$n\log_b(m)$
This rule is valid for positive values of $b,$ $m,$ and $n$ and $b\neq 1.$
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Exercise

Simplify the expression $\log_6(8)-\log_6(3)+\log_6(9)-\log_6(4)$ using the properties of logarithms.

Show Solution
Solution
The Quotient Property of Logarithms states that the following is true for logarithms. $\log_b\left(\dfrac{m}{n}\right)=\log_b(m)-\log_b(n)$ 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. $\log_6(8)-\log_6(3)+\log_6(9)-\log_6(4) \\ \Updownarrow \\ \log_6 \left(\dfrac{8}{3} \right)+\log_6 \left(\dfrac{9}{4} \right)$ 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. $\log_6 \left(\dfrac{8}{3} \right)+\log_6 \left(\dfrac{9}{4} \right) \quad \Leftrightarrow \quad \log_6 \left(\dfrac{8}{3} \cdot \dfrac{9}{4} \right)$ Let's continue simplifying this expression.
$\log_6 \left(\dfrac{8}{3} \cdot \dfrac{9}{4} \right)$
$\log_6 \left(\dfrac{8\cdot 9}{3\cdot4} \right)$
$\log_6 \left(6 \right)$
$\log_{6}({6}) = 1$
$1$
Thus, the expression $\log_6(8)-\log_6(3)+\log_6(9)-\log_6(4)$ equals $1.$
Rule

## Change of Base Formula

The Change of Base Formula allows the logarithm of an arbitrary base to be rewritten as the quotient of two logarithms with another base. $\log_c (a)= \dfrac{\log_b (a)}{\log_b (c)}$ 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.

$\log_c (a)= \frac{\log (a)}{\log (c)} \quad \text{and} \quad \log_c (a)= \frac{\ln (a)}{\ln (c)}$
Method

## Solving an Exponential Equation using Logarithms

An exponential equation can be solved by applying logarithms and using the Power Property of Logarithms. Consider the following equation. $\begin{gathered} 8^x=3 \end{gathered}$

### 1

Apply the logarithm

First, the equation is rewritten by applying a logarithm on both sides. $8^x=3 \quad \Leftrightarrow \quad \log\left( 8^x \right)=\log(3)$

### 2

Rewrite using the Power Property of Logarithms

By using the Power Property of Logarithms, powers can be rewritten into a product. $\begin{gathered} \log\left( 8^x \right)=\log(3) \\ \Updownarrow \\ x\cdot \log(8)=\log(3) \end{gathered}$

### 3

Solve the resulting equation

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 $\log(8).$ $\begin{gathered} x\cdot \log(8)=\log(3) \\ \Updownarrow \\ x=\dfrac{\log(3)}{\log(8)} \end{gathered}$ By using a calculator, an approximate value of $x$ can be calculated. Here, $x\approx 0.53.$

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Exercise

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

Show Solution
Solution
The given equation can be solved by applying the $\log_4$ on both sides. Since many calculators are limited to only the common and the natural logarithms, it is often not possible to use $\log_4.$ We can solve it anyway. Let's begin by applying the common logarithm on both sides. $4^x=25 \quad \Leftrightarrow \quad \log\left( 4^x \right)=\log(25)$ The Power Property of Logarithms gives us the relationship $\log_b\left(m^n \right)=n\cdot \log_b(m).$ With this we can rewrite the left-hand side of the equation. $\log\left( 4^x \right)=\log(25)\, \, \Leftrightarrow \, \, x\cdot \log(4)=\log(25)$ We can now solve the equation for $x$ by dividing both sides of the equation with $\log(4).$
$x\cdot \log(4)=\log(25)$
$x=\dfrac{\log(25)}{\log(4)}$
$x=2.32192\ldots$
$x \approx 2.32$
The solution to the equation is $x\approx 2.32.$
Rule

## Inverse Properties of Logarithms

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

1. $g(f(x))=\log_b \left( b^x \right) = x$
2. $f(g(x))=b^{\log _b (x)} = x$

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.

$\begin{gathered} \log_b\left(b^x\right)=x \quad \text{and} \quad b^{\log_b(x)}=x \end{gathered}$

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

\begin{aligned} \log\left(10^x\right)=x \quad \text{and}& \quad 10^{\log(x)}=x \\[0.8em] \ln\left(e^x\right)=x \quad \text{and}& \quad e^{\ln(x)}=x \end{aligned}

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.

\begin{array}{c|c} \begin{aligned} {\color{#0000FF}{\ln\left( e^{4x}\right)}}&=20\\ {\color{#0000FF}{4x}}&=20\\ x &= 5 \end{aligned} \hspace{0.35cm} & \hspace{0.35cm} \begin{aligned} {\color{#0000FF}{3^{\log_3 (5x)}}}&=10\\ {\color{#0000FF}{5x}}&=10\\ x &= 2 \end{aligned} \end{array}

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Exercise

Solve the equation $\dfrac{e^{\ln (3^4)}\cdot 10^{\log(x)}}{\log_5\left( 125 \right)}=\log(100)$ using the inverse properties of logarithms.

Show Solution
Solution
The inverse properties of logarithms tell us that a logarithm and a power with the same base undo each other. $\log_b(b^x)=x \quad \text{and} \quad b^{\log_b(x)}=x$ On the left-hand side, we can identify $\ln$ as being the logarithm with base $e$ and $\log$ 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. \begin{aligned} \dfrac{e^{\ln (3^4)}\cdot 10^{\log(x)}}{\log_5\left( 125 \right)}&=\log(100) \\ \Updownarrow \\ \dfrac{3^4\cdot x}{\log_5\left( 125 \right)}&=\log(100) \end{aligned} To simplify the remaining two logarithms, we first need to rewrite their arguments. $\small \dfrac{3^4\cdot x}{\log_5\left( 125 \right)}=\log(100) \, \, \Leftrightarrow \, \, \dfrac{3^4\cdot x}{\log_5\left( 5^3 \right)}=\log\left( 10^2 \right)$ Again, we can use the the inverse properties of logarithms to simplify the equation. $\dfrac{3^4\cdot x}{\log_5\left( 5^3 \right)}=\log\left( 10^2 \right) \, \, \Leftrightarrow \, \, \dfrac{3^4\cdot x}{3}=2$ We now need to isolate $x$ on the left-hand side to find the solution.
$\dfrac{3^4\cdot x}{3}=2$
$3^3\cdot x=2$
$x=\dfrac{2}{3^3}$
$x=\dfrac{2}{27}$
We have solved the equation, and its solution is $\frac{2}{27}.$