3.6 Properties of Logarithms

Before we learn about logarithmic functions, we will outline the properties of logarithms like we did for exponents. There are nine basic properties of logarithms. Let $u$ $u$ and $v$ $v$ be real numbers with $u > 0$ $u > 0$ and $v > 0$ $v > 0$ . Moreover, let $a$ $a$ and $b$ $b$ be real numbers with $a > 0$ $a > 0$ , $a \neq 1$ $a \neq 1$ , $b > 0$ $b > 0$ , and $b \neq 1$ $b \neq 1$ .

1. Definition of a Logarithm

The expression

$v = {log}_{a} u$ $v = \log_{a} u$

is read as “ $v$ $v$ is the logarithm base $a$ $a$ of $u$ $u$ ,” if and only if $u = a^{v}$ $u = a^{v}$ . The word logarithm is often shortened to log. Note that $u > 0$ $u > 0$ , so u has to be a positive number.

Here are a few examples:

${log}_{5} 25 = 2 ⟺ 25 = 5^{2}$ $\log_{5} 25 = 2 ⟺ 25 = 5^{2}$

${log}_{3} \frac{1}{3} = - 1 ⟺ \frac{1}{3} = 3^{- 1}$ $\log_{3} \frac{1}{3} = - 1 ⟺ \frac{1}{3} = 3^{- 1}$

${log}_{3} \frac{1}{81} = - 4 ⟺ \frac{1}{9} = 3^{- 4}$ $\log_{3} \frac{1}{81} = - 4 ⟺ \frac{1}{9} = 3^{- 4}$

${log}_{2} \sqrt{2} = \frac{1}{2} ⟺ \sqrt{2} = 2^{\frac{1}{2}}$ $\log_{2} \sqrt{2} = \frac{1}{2} ⟺ \sqrt{2} = 2^{\frac{1}{2}}$

${log}_{4} 4 = 1 ⟺ 4 = 4^{1}$ $\log_{4} 4 = 1 ⟺ 4 = 4^{1}$

${log}_{7} 1 = 0 ⟺ 1 = 7^{0}$ $\log_{7} 1 = 0 ⟺ 1 = 7^{0}$

2. Common and Natural Logarithms

Logarithms to base 10 are called common logarithms. For simplicity, ${log}_{10} u$ ${log}_{10} u$ is abbreviated as $log u$ $log u$ , without writing the base. The most commonly used type of logarithms are natural logarithms, that is, logarithms to base $e$ $e$ . The conventional notation of natural logarithms is $ln u$ $\ln u$ instead of ${log}_{e} u$ $\log_{e} u$ .

3. Trivial Identities of Logarithms

There are two trivial identities of logarithms. The first is that a logarithm to the base $a$ $a$ of 1 is 0. Mathematically,

${log}_{a} 1 = 0$ $\log_{a} 1 = 0$

The second is that a logarithm to the base $a$ $a$ of $a$ $a$ is 1. Mathematically,

${log}_{a} a = 1$ $\log_{a} a = 1$

4. Power Rule of Logarithms

A logarithm to the base $a$ $a$ of some exponential number $u^{r}$ $u^{r}$ is the product of $r$ $r$ and the logarithm to the base $a$ $a$ of $u$ $u$ .

${log}_{a} (u^{r}) = r \cdot {log}_{a} u$ $\log_{a} (u^{r}) = r \cdot \log_{a} u$

Note that ${log}_{a} (u^{r}) \neq ({log}_{a} u)^{r}$ $\log_{a} (u^{r}) \neq (\log_{a} u)^{r}$ when $u \neq 1$ $u \neq 1$ and $u \neq a$ $u \neq a$ .

Here are a few examples.

${log}_{2} (5^{6}) = 6 {log}_{2} 5$ $\log_{2} (5^{6}) = 6 \log_{2} 5$

${log}_{5} (13^{- 3}) = - 3 {log}_{5} 13$ $\log_{5} (13^{- 3}) = - 3 \log_{5} 13$

${log}_{3} \sqrt{5} = {log}_{3} 5^{\frac{1}{2}} = \frac{1}{2} {log}_{3} 5$ $\log_{3} \sqrt{5} = \log_{3} 5^{\frac{1}{2}} = \frac{1}{2} \log_{3} 5$

5. Canceling Logarithms

Given Property 4 above, a logarithm to the base $a$ $a$ of some exponent power of $a$ $a$ , $a^{r}$ $a^{r}$ , is $r$ $r$ .

${log}_{a} (a^{u}) = u$ $\log_{a} (a^{u}) = u$

This is because ${log}_{a} a = 1$ $\log_{a} a = 1$ :

${log}_{a} (a^{u}) = u \cdot {log}_{a} a$ $\log_{a} (a^{u}) = u \cdot \log_{a} a$
$= u \cdot 1$ $= u \cdot 1$
$= u$ $= u$

Moreover,

$a^{{log}_{a} u} = u$ $a^{\log_{a} u} = u$

This is because, based on the definition of a logarithm,

$a^{{log}_{a} u} = u ⟺ {log}_{a} u = l o g_{a} u$ $a^{\log_{a} u} = u ⟺ \log_{a} u = l o g_{a} u$

6. Logarithm of a Product

The logarithm of a product is the sum of the logarithms of the factors.

${log}_{a} (u \cdot v) = {log}_{a} u + {log}_{a} v$ $\log_{a} (u \cdot v) = \log_{a} u + \log_{a} v$

7. Logarithm of a Quotient

The logarithm of a quotient is the logarithm of the numerator minus the logarithm of the denominator.

${log}_{a} (\frac{u}{v}) = {log}_{a} u - {log}_{a} v$ $\log_{a} (\frac{u}{v}) = \log_{a} u - \log_{a} v$

8. Change of Base

The base of a logarithm can be changed from $a$ $a$ to $b$ $b$ with the following formula:

${log}_{a} u = \frac{{log}_{b} u}{{log}_{b} a}$ $\log_{a} u = \frac{\log_{b} u}{\log_{b} a}$

9. Taking Logarithms on Both Sides

Suppose that

$u = v$ $u = v$

Then the equation holds that if both sides of equation take a logarithm of the same base,

${log}_{b} u = {log}_{b} v$ $\log_{b} u = \log_{b} v$

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