# 6.5 Logarithmic properties  (Page 3/10)

 Page 3 / 10

## Using the quotient rule for logarithms

Expand $\text{\hspace{0.17em}}{\mathrm{log}}_{2}\left(\frac{15x\left(x-1\right)}{\left(3x+4\right)\left(2-x\right)}\right).$

First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.

${\mathrm{log}}_{2}\left(\frac{15x\left(x-1\right)}{\left(3x+4\right)\left(2-x\right)}\right)={\mathrm{log}}_{2}\left(15x\left(x-1\right)\right)-{\mathrm{log}}_{2}\left(\left(3x+4\right)\left(2-x\right)\right)$

Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5.

Expand $\text{\hspace{0.17em}}{\mathrm{log}}_{3}\left(\frac{7{x}^{2}+21x}{7x\left(x-1\right)\left(x-2\right)}\right).$

${\mathrm{log}}_{3}\left(x+3\right)-{\mathrm{log}}_{3}\left(x-1\right)-{\mathrm{log}}_{3}\left(x-2\right)$

## Using the power rule for logarithms

We’ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as $\text{\hspace{0.17em}}{x}^{2}?\text{\hspace{0.17em}}$ One method is as follows:

$\begin{array}{ll}{\mathrm{log}}_{b}\left({x}^{2}\right)\hfill & ={\mathrm{log}}_{b}\left(x\cdot x\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}x+{\mathrm{log}}_{b}x\hfill \\ \hfill & =2{\mathrm{log}}_{b}x\hfill \end{array}$

Notice that we used the product rule for logarithms    to find a solution for the example above. By doing so, we have derived the power rule for logarithms , which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,

$\begin{array}{lll}100={10}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill & \sqrt{3}={3}^{\frac{1}{2}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hfill & \frac{1}{e}={e}^{-1}\hfill \end{array}$

## The power rule for logarithms

The power rule for logarithms    can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.

${\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M$

Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.

1. Express the argument as a power, if needed.
2. Write the equivalent expression by multiplying the exponent times the logarithm of the base.

## Expanding a logarithm with powers

Expand $\text{\hspace{0.17em}}{\mathrm{log}}_{2}{x}^{5}.$

The argument is already written as a power, so we identify the exponent, 5, and the base, $\text{\hspace{0.17em}}x,$ and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

${\mathrm{log}}_{2}\left({x}^{5}\right)=5{\mathrm{log}}_{2}x$

Expand $\text{\hspace{0.17em}}\mathrm{ln}{x}^{2}.\text{\hspace{0.17em}}$

$2\mathrm{ln}x$

## Rewriting an expression as a power before using the power rule

Expand $\text{\hspace{0.17em}}{\mathrm{log}}_{3}\left(25\right)\text{\hspace{0.17em}}$ using the power rule for logs.

Expressing the argument as a power, we get $\text{\hspace{0.17em}}{\mathrm{log}}_{3}\left(25\right)={\mathrm{log}}_{3}\left({5}^{2}\right).$

Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

${\mathrm{log}}_{3}\left({5}^{2}\right)=2{\mathrm{log}}_{3}\left(5\right)$

Expand $\text{\hspace{0.17em}}\mathrm{ln}\left(\frac{1}{{x}^{2}}\right).$

$-2\mathrm{ln}\left(x\right)$

## Using the power rule in reverse

Rewrite $\text{\hspace{0.17em}}4\mathrm{ln}\left(x\right)\text{\hspace{0.17em}}$ using the power rule for logs to a single logarithm with a leading coefficient of 1.

Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression $\text{\hspace{0.17em}}4\mathrm{ln}\left(x\right),$ we identify the factor, 4, as the exponent and the argument, $\text{\hspace{0.17em}}x,$ as the base, and rewrite the product as a logarithm of a power: $\text{\hspace{0.17em}}4\mathrm{ln}\left(x\right)=\mathrm{ln}\left({x}^{4}\right).\text{\hspace{0.17em}}$

Rewrite $\text{\hspace{0.17em}}2{\mathrm{log}}_{3}4\text{\hspace{0.17em}}$ using the power rule for logs to a single logarithm with a leading coefficient of 1.

${\mathrm{log}}_{3}16$

## Expanding logarithmic expressions

Taken together, the product rule, quotient rule, and power rule are often called “laws of logs.” Sometimes we apply more than one rule in order to simplify an expression. For example:

x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
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