# 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:

An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
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Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
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salma
Commplementary angles
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Tamia
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Uday
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salma
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×