4.4 Properties of Logarithms

Introduction

Because logarithms are actually exponents, they have several properties that can be derived from the laws of exponents. Here are the laws we will need at present.

Each of these laws corresponds to one of three properties of logarithms.

We will consider proofs of the three properties of logarithms in the Homework problems. For now, study the examples below, keeping in mind that a logarithm is an exponent.

Note

Property (1):
$\begin{aligned} \log_{2} (32) = \log_{2} (4 \cdot 8) & = \log_{2} 4 + \log_{2} 8 because 2^{5} = 2^{2} \cdot 2^{3} \\ 5 & = 2 + 3 32 = 4 \cdot 8 \end{aligned}$
Property (2):
$\begin{aligned} \log_{2} (8) = \log_{2} (\frac{16}{2}) & = \log_{2} (16) - \log_{2} (2) because 2^{3} = \frac{2^{4}}{2^{1}} \\ 3 & = 4 - 1 8 = \frac{16}{2} \end{aligned}$
Property (3): Property (3):
$\begin{aligned} \log_{2} 64 = \log_{2} (4)^{3} & = 3 \log_{2} (4) because {(2^{2})}^{3} = 2^{6} \\ 6 & = 3 \cdot 2 (4)^{3} = 64 \end{aligned}$

Which statement is true?

_____

$\log (8 y) = \log (8) + \log (y)$

Which statement is true?

$\log (8 + y) = \log (8) + \log (y)$
$\log (8 y) = \log (8) + \log (y)$
$\log (8 y) = \log (8) \cdot \log (y)$
$\log (\frac{8}{y}) = \frac{\log (8)}{\log (y)}$

Using the Properties of Logarithms

Of course, these properties are useful not so much for computing logs but rather for simplifying expressions that contain variables. We will use them to solve exponential equations. But first, we will practice applying the properties. In the following Example, we rewrite one log in terms of simpler logs.

Example 1

Simplify $\log_{b} (\sqrt{x y})$ .

First, we write $\sqrt{x y}$ using a fractional exponent:

$\log_{b} (x y) = \log_{b} (x y)^{1 / 2}$

Then we apply Property (3) to rewrite the exponent as a coefficient:

$\log_{b} (x y)^{1 / 2} = \frac{1}{2} \log_{b} (x y)$

Finally, by Property (1) we write the log of a product as a sum of logs:

$\frac{1}{2} (\log_{b} (x y)) = \frac{1}{2} (\log_{b} (x) + \log_{b} (y))$

Thus, $\log_{b} (\sqrt{x y}) = \frac{1}{2} (\log_{b} (x) + \log_{b} (y))$ .

Simplify $\log_{b} (\frac{x}{y^{2}})$ .

Answer: $\log_{b}$ _____ $+$ _____ $\cdot \log_{b}$ _____

$log sub b ⁡ x minus 2 log sub b ⁡ y$

Simplify $\log_{b} (\frac{x}{y^{2}})$ .

$log sub b ⁡ open parenthesis x close parenthesis minus 2 log sub b ⁡ open parenthesis y close parenthesis$

Caution

Be careful when using the properties of logarithms! Compare the statements below:

$\log_{b} (2 x) = \log_{b} (2) + \log_{b} (x) by Property 1,$
but
$\log_{b} (2 + x) \neq \log_{b} (2) + \log_{b} (x)$
$\log_{b} (\frac{x}{5}) = \log_{b} (x) - \log_{b} (5) by Property 2,$
but
$\log_{b} (\frac{x}{5}) \neq \frac{\log_{b} (x)}{\log_{b} (5)}$

We can also use the properties of logarithms to combine sums and differences of logarithms into one logarithm.

Example 2

Express $3 open parenthesis log sub b ⁡ open parenthesis x close parenthesis minus log sub b ⁡ open parenthesis y close parenthesis close parenthesis$ as a single logarithm with a coefficient of $1$ .

We begin by applying Property (2) to combine the logs.

$3 (\log_{b} (x) - \log_{b} (y)) = 3 \log_{b} (\frac{x}{y})$

Then, using Property (3), we replace the coefficient $3$ by an exponent $3$ .

$3 \log_{b} (\frac{x}{y}) = \log_{b} ({(\frac{x}{y})}^{3})$

Express $2 \log_{b} x + 4 \log_{b} (x + 3)$ as a single logarithm with a coefficient of $1$ .

Answer: $\log_{b}$ _____

$\log_{b} x^{2} (x + 3)^{4}$

Express $2 \log_{b} (x) + 4 \log_{b} (x + 3)$ as a single logarithm with a coefficient of $1$ .

$\log_{b} (x^{2} (x + 3)^{4})$

Which expression is equivalent to $3 log sub 2 ⁡ open parenthesis H close parenthesis minus log sub 2 ⁡ open parenthesis T close parenthesis$ ?

_____

$\log_{2} (\frac{H^{3}}{T})$

Which expression is equivalent to $3 log sub 2 ⁡ open parenthesis H close parenthesis minus log sub 2 ⁡ open parenthesis T close parenthesis$ ?

$3 log sub 2 ⁡ open parenthesis H minus T close parenthesis$
$log sub 2 ⁡ open parenthesis H cubed minus T close parenthesis$
$\log_{2} (\frac{H^{3}}{T})$
$\log_{2} ({(\frac{H}{T})}^{3})$

Solving Exponential Equations

By using Property (3), we can now solve exponential equations in which the base is not $10$ . For example, to solve the equation

$5^{x} = 7$

we could rewrite the equation in logarithmic form to obtain the exact solution

$x equals log sub 5 ⁡ open parenthesis 7 close parenthesis$

However, we cannot evaluate $log sub 5 ⁡ open parenthesis 7 close parenthesis$ ; there is no log base $5$ button on the calculator. If we want a decimal approximation for the solution, we begin by taking the base $10$ logarithm of both sides, even though the base of the power is not $10$ . This gives us

$\log_{10} (5^{x}) = \log_{10} (7)$

Then we use Property (3) to rewrite the left side as

$x log sub 10 ⁡ open parenthesis 5 close parenthesis equals log sub 10 ⁡ open parenthesis 7 close parenthesis$

Note how using Property (3) allows us to solve the equation: The variable, $x$ , is no longer in the exponent, and it is multiplied by a constant, $log sub 10 ⁡ open parenthesis 5 close parenthesis$ . To finish the solution, we divide both sides by $log sub 10 ⁡ open parenthesis 5 close parenthesis$ to get

$x = \frac{\log_{10} (7)}{\log_{10} (5)}$

On your calculator, enter the sequence

LOG $7$ ) ÷ LOG $5$ ) ENTER

to find that $x approximately equals 1.2091$ .

Caution

Do not confuse the expression $\frac{\log_{10} (7)}{\log_{10} (5)}$ with $\log_{10} (\frac{7}{5})$ ; they are not the same!

Property (2) allows us to simplify $\log (\frac{x}{y})$ , but not $\frac{\log (x)}{\log (y)}$ . We cannot rewrite $\frac{\log_{10} (7)}{\log_{10} (5)}$ , so we must evaluate it as $open parenthesis log ⁡ 7 close parenthesis / open parenthesis log ⁡ open parenthesis 5 close parenthesis close parenthesis$ . You can check on your calculator that

$\frac{\log_{10} (7)}{\log_{10} (5)} \neq \log_{10} (\frac{7}{5}) = \log_{10} (1.4)$

How can we solve $4^{x} = 60$ ?

_____

Take log base 10 of both sides.

How can we solve $4^{x} = 60$ ?

Divide both sides by 4.
Take log base 10 of both sides.
Take the fourth root of 60.
This equation has no solution.

Example 3

Solve $1640 = 80 \cdot 6^{0.03 x}$

First we divide both sides by $80$ to obtain

$20.5 = 6^{0.03 x}$

Next, we take the base $10$ logarithm of both sides of the equation and use Property (3) of logarithms to get

$\log_{10} (20.5) = \log_{10} (6^{0.03 x}) = 0.03 x \log_{10} (6)$

On the right side of the equation, $x$ is multiplied by two constants, $0.03$ and $log sub 10 ⁡ open parenthesis 6 close parenthesis$ . So, to solve for $x$ we must divide both sides of the equation by $0.03 log sub 10 ⁡ open parenthesis 6 close parenthesis$ . We use a calculator to evaluate the answer:

$x = \frac{\log_{10} (20.5)}{0.03 \log_{10} (6)} \approx 56.19$

(On your calculator, remember to enclose the denominator, $0.03 log sub 10 ⁡ open parenthesis 6 close parenthesis$ , in parentheses.)

We summarize our method for solving exponential equations as follows.

Solve $5 (1.2)^{2.5 x} = 77$

$x =$ _____

$Divide both sides by 5.$

$Take the log of both sides.$

$Apply Property (3) to simplify the left side.$

$Solve for x .$

$x = \frac{\log (15.4)}{2.5 \log (1.2)} \approx 5.999$

Solve $5 (1.2)^{2.5 x} = 77$

Divide both sides by 5, then take the log of both sides. Simplify the left side and solve for $x$ to find $x = \frac{\log (15.4)}{2.5 \log (1.2)} \approx 5.999$

Explain why the distributive law does not apply to the expression $\log (a + b)$ .

_____

Explain why the distributive law does not apply to the expression $\log (a + b)$ .

Applications

By using the properties of logarithms, we can now solve equations that arise in exponential growth and decay models, no matter what base the exponential function uses.

Example 4

The population of Silicon City was $6500$ in $1990$ and has been tripling every $12$ years. When will the population reach $75, 000$ ?

The population of Silicon City grows according to the formula

$P (t) = 6500 \cdot 3^{t / 12}$

where $t$ is the number of years after $1990$ . We want to find the value of $t$ for which $P open parenthesis t close parenthesis equals 75, 000$ ; that is, we want to solve the equation

$\begin{aligned} 6500 \cdot 3^{t / 12} & = 75, 000 & Divide both sides by 6500. \\ 3^{t / 12} & = \frac{150}{13} \end{aligned}$

Now we take the base $10$ logarithm of both sides and solve for $t$ .

$\begin{aligned} \log_{10} (3^{t / 12}) & = \log_{10} (\frac{150}{13}) & Apply Property (3). \\ \frac{t}{12} \log_{10} (3) & = \log_{10} (\frac{150}{13}) & Divide by \log_{10} (3); multiply by 12. \\ t & = \frac{12 \log_{10} (\frac{150}{13})}{\log_{10} (3)} \\ \approx 26.71 \end{aligned}$

The population of Silicon City will reach $75, 000$ about $27$ years after $1990$ , or in $2017$ .

Traffic on U.S. highways is growing by 2.7% per year. (Source: Time, Jan. 25, 1999)

Write a formula for the volume, $V$ , of traffic as a function of time, using $V sub 0$ for the current volume. [Note: Enter "V0" to get $V sub 0$ .]
$V =$ _____
How long will it take the volume of traffic to double? Hint: Find the value of $t$ that gives $V equals 2 V sub 0$ .
Answer: about _____ years

$V (t) = V_{0} (1.027)^{t}$
about $26$ years

Traffic on U.S. highways is growing by 2.7% per year. (Source: Time, Jan. 25, 1999)

Write a formula for the volume, $V$ , of traffic as a function of time, using $V sub 0$ for the current volume.
How long will it take the volume of traffic to double? Hint: Find the value of $t$ that gives $V equals 2 V sub 0$ .

$V (t) = V_{0} (1.027)^{t}$
about $26$ years

How does Property (3) of logarithms help us solve exponential equations?

_____

How does Property (3) of logarithms help us solve exponential equations?

Compound Interest

The amount of money in an account that earns interest compounded annually grows exponentially according to the formula

$A (t) = P (1 + r)^{t}$

(See Exponential Growth and Decay to review compound interest.) Many accounts compound interest more frequently than once a year. If the interest is compounded $n$ times per year, then in $t$ years there will be $n t$ compounding periods, and in each period the account earns interest at a rate of $\frac{r}{n}$ . The amount accumulated is given by a generalization of our earlier formula.

Note

Compound Interest

The amount $A (t)$ accumulated (principal plus interest) in an account bearing interest compounded $n$ times annually is

$A (t) = P {(1 + \frac{r}{n})}^{n t}$

where

		$P$	is the principal invested,
		$r$	is the interest rate,
		$t$	is the time period, in years.

Example 5

Rashad deposited $ $1000$ in an account that pays $4$ % interest. Calculate the amount in his account after $5$ years if the interest is compounded

semiannually
quarterly
monthly

Semiannually means "twice a year,"" so we use the formula for compound interest with $P equals 1000$ , $r equals 0.04$ , $n equals 2$ , and $t equals 5$ .
$\begin{aligned} A (5) & = 1000 {(1 + \frac{0.04}{2})}^{2 (5)} \\ = 1000 (1.02)^{10} = 1218.99 \end{aligned}$
If interest is compounded semiannually, the balance in the account after $5$ years is $ $1218.99$ .
Quarterly means "4 times a year,"" so we use the formula for compound interest with $P equals 1000$ , $r equals 0.04$ , $n equals 4$ , and $t equals 5$ .
$\begin{aligned} A (5) & = 1000 {(1 + \frac{0.04}{4})}^{4 (5)} \\ = 1000 (1.01)^{20} = 1220.19 \end{aligned}$
If interest is compounded quarterly, the balance in the account after $5$ years is $ $1220.19$ .
There are $12$ months in a year, so we use the formula for compound interest with $P equals 1000$ , $r equals 0.04$ , $n equals 12$ , and $t equals 5$ .
$\begin{aligned} A (5) & = 1000 {(1 + \frac{0.04}{12})}^{12 (5)} \\ = 1000 (1.003)^{60} = 1221.00 \end{aligned}$
If interest is compounded monthly, the balance in the account after $5$ years is $ $1221$ .

Which formula gives the amount when interest is compounded quarterly?

_____

$P {(1 + \frac{r}{4})}^{4 t}$

Which formula gives the amount when interest is compounded quarterly?

$P (1 + 4 r)^{t}$
$P (1 + 4 r)^{4 t}$
$P {(1 + \frac{r}{4})}^{4 t}$
$P {(1 + \frac{r}{4})}^{t / 4}$

Calculate the amount in Rashad's account after 5 years if the interest is compounded daily. (See Example. There are 365 days in a year.)

Answer: $_____

$1221.39

Calculate the amount in Rashad's account after 5 years if the interest is compounded daily. (See Example. There are 365 days in a year.)

$1221.39

Solving Formulas

The techniques for solving exponential equations can also be used to solve formulas involving exponential expressions for one variable in terms of the others.

Example 6

Solve $2 C = C b^{k t}$ for $t$ . (Assume that $C$ and $k not equal to 0$ .)

First, we divide both sides by $C$ to isolate the power.

$b^{k t} = 2$

Next, we take the log base $10$ of both sides.

$\begin{aligned} \log (b^{k t}) & = \log 2 & Apply Log Property (3). \\ k t \log (b) & = \log 2 \end{aligned}$

Finally, we divide both sides by $k \log (b)$ to solve for $t$ .

$t = \frac{\log (2)}{k \log (b)}$

Solve $A = P (1 + r)^{t}$ for $t$ .

$t =$ _____

$t = \frac{\log (A / P)}{\log (1 + r)}$

Solve $A = P (1 + r)^{t}$ for $t$ .

$t = \frac{\log (A / P)}{\log (1 + r)}$

Section Summary

Vocabulary

Look up the definitions of new terms in the Glossary.

Compounding period

CONCEPTS

Note
Properties of Logarithms
If $x$ , $y$ , $b greater than 0$ , and $b not equal to 1$ , then
1. $\log_{b} (x y) = \log_{b} (x) + \log_{b} (y)$
2. $\log_{(b)} \frac{x}{y} = \log_{b} (x) - \log_{b} (y)$
3. $\log_{b} (x^{k}) = k \log_{b} (x)$
We can use the properties of logarithms to solve exponential equations with any base.
Note
Steps for Solving Exponential Equations
1. Isolate the power on one side of the equation.
2. Take the log base $10$ of both sides.
3. Simplify by applying Log Property (3).
4. Solve for the variable.
The amount in an account earning interest compounded $n$ times per year is an exponential function of time.
Note
Compound Interest
The amount $A (t)$ accumulated (principal plus interest) in an account bearing interest compounded $n$ times annually is
$A (t) = P {(1 + \frac{r}{n})}^{n t}$
where
$P$ is the principal invested,
$r$ is the interest rate,
$t$ is the time period, in years.

STUDY QUESTIONS

The properties of logs are really another form of which familiar laws?
Which log property allows us to solve an exponential equation whose base is not $10$ ?
Explain why $12 \cdot 10^{3 x}$ is not the same as $120 to the power 3 x$ .
Which of the following expressions are equivalent?
$\log (\frac{x}{4}) \frac{\log (x)}{\log (4)} \log (x - 4) \log (x) - \log (4)$
Which of the following expressions are equivalent?
$\log (x + 2) \log (x) + \log (2) \log (2 x) (\log (2)) (\log (x))$
Which of the following expressions are equivalent?
$log ⁡ open parenthesis x cubed close parenthesis open parenthesis log ⁡ open parenthesis 3 close parenthesis close parenthesis open parenthesis log ⁡ open parenthesis x close parenthesis close parenthesis 3 log ⁡ open parenthesis x close parenthesis log ⁡ open parenthesis 3 to the power x close parenthesis$

SKILLS

Practice each skill in the Homework problems listed.

Use the properties of logarithms to simplify expressions: #1–24, #45–52
Solve exponential equations using logs base 10: #25–36
Solve problems about exponential models: #37–44
Solve problems about compound interest: #53–58
Solve formulas involving exponential expressions: #59–64

Homework 4.4

Simplify $10^{2} \cdot 10^{6}$ .
Compute $log ⁡ open parenthesis 10 squared close parenthesis$ , $log ⁡ open parenthesis 10 to the power 6 close parenthesis$ , and $\log (10^{2} \cdot 10^{6})$ . How are they related?

$10 to the power 8$
$2$ ; $6$ ; $8$ ; $2 + 6 = 8$

Simplify $\frac{10^{9}}{10^{6}}$ .
Compute $log ⁡ open parenthesis 10 to the power 9 close parenthesis$ , $log ⁡ open parenthesis 10 to the power 6 close parenthesis$ , and $\log (\frac{10^{9}}{10^{6}})$ . How are they related?

Simplify $\frac{b^{8}}{b^{5}}$ .
Compute $log sub b ⁡ open parenthesis b to the power 8 close parenthesis$ , $log sub b ⁡ open parenthesis b to the power 5 close parenthesis$ , and $\log_{b} (\frac{b^{8}}{b^{5}})$ . How are they related?

$b^{3}$
$8$ ; $5$ ; $3$ ; $8 - 5 = 3$

Simplify $b^{4} \cdot b^{3}$ .
Compute $log sub b ⁡ open parenthesis b to the power 4 close parenthesis$ , $\log_{b} (b^{3})$ , and $\log_{b} (b^{4} \cdot b^{3})$ . How are they related?

Simplify $open parenthesis 10 cubed close parenthesis to the power 5$ .
Compute $log ⁡ open parenthesis 10 cubed close parenthesis to the power 5$ and $log ⁡ open parenthesis 10 cubed close parenthesis$ . How are they related?

$10 to the power 15$
$15$ ; $3$ ; $15 = 3 \cdot 5$

Simplify $open parenthesis b squared close parenthesis to the power 6$ .
Compute $log sub b ⁡ open parenthesis open parenthesis b squared close parenthesis to the power 6 close parenthesis$ and $\log_{b} (b^{2})$ . How are they related?

For Problems 7–14, use the properties of logarithms to expand each expression in terms of simpler logarithms. Assume that all variable expressions denote positive numbers.

$log sub b ⁡ open parenthesis 2 x close parenthesis$
$\log_{b} (\frac{x}{2})$

$log sub b ⁡ open parenthesis 2 close parenthesis plus log sub b ⁡ open parenthesis x close parenthesis$
$log sub b ⁡ open parenthesis 2 close parenthesis minus log sub b ⁡ open parenthesis x close parenthesis$

$\log_{b} (\frac{2 x}{x - 2})$
$log sub b ⁡ open parenthesis x open parenthesis 2 x plus 3 close parenthesis close parenthesis$

$log sub 3 ⁡ open parenthesis 3 x to the power 4 close parenthesis$
$log sub 5 ⁡ open parenthesis 1.1 to the power 1 / t close parenthesis$

$1 plus 4 log sub 3 ⁡ x$
$\frac{1}{t} \log_{5} (1.1)$

$log sub b ⁡ open parenthesis open parenthesis 4 b close parenthesis to the power t close parenthesis$
$\log_{2} (5 \cdot 2^{x})$

$\log_{b} (\sqrt{b x})$
$\log_{3} (\sqrt[3]{x^{2} + 1})$

$\frac{1}{2} + \frac{1}{2} \log_{b} (x)$
$\frac{1}{3} \log_{3} (x^{2} + 1)$

$\log_{10} (\sqrt{\frac{2 L}{R^{2}}})$
$\log_{10} (2 π \sqrt{\frac{l}{g}})$

$\log (P_{0} (1 - m)^{t})$
$\log_{4} {(1 + \frac{r}{4})}^{4 t}$

$\log P_{0} + t \log (1 - m)$
$4 t [\log_{4} (4 + r) - 1]$

$\log_{3} (\frac{a^{2} - 2}{a^{5}})$
$\log (\frac{a^{3} b^{2}}{(a + b)^{3 / 2}})$

For Problems 15–20, combine into one logarithm and simplify. Assume all expressions are defined.

$log sub b ⁡ open parenthesis 8 close parenthesis minus log sub b ⁡ open parenthesis 2 close parenthesis$
$2 log sub 4 ⁡ open parenthesis x close parenthesis plus 3 log sub 4 ⁡ open parenthesis y close parenthesis$

$log sub b ⁡ open parenthesis 4 close parenthesis$
$log sub 4 ⁡ open parenthesis x squared y cubed close parenthesis$

$log sub b ⁡ open parenthesis 5 close parenthesis plus log sub b ⁡ open parenthesis 2 close parenthesis$
$\frac{1}{4} \log_{5} (x) - \frac{3}{4} \log_{5} (y)$

$\log (2 x) + 2 \log (x) - \log (\sqrt{x})$
$\log (t^{2} - 16) - \log (t + 4)$

$log ⁡ open parenthesis 2 x to the power 5 / 2 close parenthesis$
$log ⁡ open parenthesis t minus 4 close parenthesis$

$\log (x^{2}) + \log (x^{3}) - 5 \log (x)$
$\log (x^{2} - x) - \log (\sqrt{x^{3}})$

$3 minus 3 log ⁡ open parenthesis 30 close parenthesis$
$\frac{1}{3} \log_{6} (8 w^{6})$

$\log \frac{1}{27}$
$log sub 6 ⁡ open parenthesis 2 w squared close parenthesis$

$2 minus log sub 4 ⁡ open parenthesis 16 z squared close parenthesis$
$1 minus 2 log sub 3 ⁡ open parenthesis x close parenthesis$

For Problems 21–24, use the three logs below to find the value of each expression.

$\log_{b} (2) = 0.6931, \log_{b} (3) = 1.0986, \log_{b} (5) = 1.6094$

(Hint: For example, $\log_{b} (15) = \log_{b} (3) + \log_{b} (5)$ .)

$log sub b ⁡ open parenthesis 6 close parenthesis$
$\log_{b} (\frac{2}{5})$

$1.7917$
$minus 0.9163$

$log sub b ⁡ open parenthesis 10 close parenthesis$
$\log_{b} (\frac{3}{2})$

$log sub b ⁡ open parenthesis 9 close parenthesis$
$log sub b ⁡ open parenthesis the square root of 50 close parenthesis$

$2.1972$
$1.9560$

$log sub b ⁡ open parenthesis 25 close parenthesis$
$log sub b ⁡ open parenthesis 75 close parenthesis$

For Problems 25–36, solve the equation by using logarithms base $10$ . Round answers to four decimal places.

$2^{x} = 7$

$2.8074$

$3^{x} = 4$

$3^{x + 1} = 8$

$0.8928$

$2^{x - 1} = 9$

$4^{x^{2}} = 15$

$\pm 1.3977$

$3^{x^{2}} = 21$

${4.26}^{- x} = 10.3$

$minus 1.6092$

${2.13}^{- x} = 8.1$

$25 \cdot 3^{2.1 x} = 47$

$0.2736$

$12 \cdot 5^{1.5 x} = 85$

$3600 = 20 \cdot 8^{- 0.2 x}$

$minus 12.4864$

$0.06 = 50 \cdot 4^{- 0.6 x}$

If raw meat is allowed to thaw at $50 degrees$ F, Salmonella grows at a rate of $9 %$ per hour.

Write a formula for the amount of Salmonella present after $t$ hours, if the initial amount is $S sub 0$ .
Health officials advise that the amount of Salmonella initially present in meat should not be allowed to increase by more than $50 %$ . How long can meat be left to thaw at $50 degrees$ F?

$S (t) = S_{0} (1.09)^{t}$
$4.7$ hours

Starting in 1998, the demand for electricity in Ireland grew at a rate of $5.8 %$ per year. In 1998, $20, 500$ gigawatts were used. (Source: Electricity Supply Board of Ireland)

Write a formula for electricity demand in Ireland as a function of time.
If demand continues to grow at the same rate, when would it reach $30, 000$ gigawatts?

The concentration of a certain drug injected into the bloodstream decreases by $20 %$ each hour as the drug is eliminated from the body. The initial dose creates a concentration of $0.7$ milligrams per milliliter.

Write a function for the concentration of the drug as a function of time.
The minimum effective concentration of the drug is $0.4$ milligrams per milliliter. When should the second dose be administered?
Verify your answer with a graph.

$C (t) = 0.7 (0.80)^{t}$
After $2.5$ hours

A small pond is tested for pollution, and the concentration of toxic chemicals is found to be $80$ parts per million. Clean water enters the pond from a stream, mixes with the polluted water, then leaves the pond so that the pollution level is reduced by $10 %$ each month.

Write a function for the concentration of toxic chemicals as a function of time.
How long will it be before the concentration of toxic chemicals reaches a safe level of $25$ parts per million?
Verify your answer with a graph.

According to the National Council of Churches, the fastest growing denomination in the United States in $2004$ was the Jehovah's Witnesses, with an annual growth rate of $1.82 %$ .

The Jehovah's Witnesses had $1, 041, 000$ members in $2004$ . Write a formula for the membership in the Jehovah's Witnesses as a function of time, assuming that the church continues to grow at the same rate.
When will the Jehovah's Witnesses have $2, 000, 000$ members?

$J (t) = 1, 041, 000 \cdot {1.0182}^{t}$
In $2040$

In $2004$ , the Presbyterian Church had $3, 241, 000$ members, but membership was declining by $4.87 %$ annually.

Write a formula for the membership in the Presbyterian Church as a function of time, assuming that the membership continues to decline at the same rate.
When will the Presbyterian Church have $2, 000, 000$ members?

Sodium-24 is a radioactive isotope that is used in diagnosing circulatory disease. It decays into stable isotopes of sodium at a rate of $4.73 %$ per hour.

Technicians inject a quantity of sodium-24 into a patient's bloodstream. Write a formula for the amount of sodium-24 present in the bloodstream as a function of time.
How long will it take for $75 %$ of the isotope to decay?

$S (t) = S_{0} \cdot {0.9527}^{t}$
$28.61$ hours

The population of Afghanistan is growing at $2.6 %$ per year.

Write a formula for the population of Afghanistan as a function of time.
In 2005, the population of Afghanistan was $29.9$ million. At the given rate of growth, how long would it take the population to reach $40$ million?

For Problems 46–52, evaluate each expression. Which (if any) are equal?

$log sub 2 ⁡ open parenthesis 4 times 8 close parenthesis$
$log sub 2 ⁡ open parenthesis 4 close parenthesis times log sub 2 ⁡ open parenthesis 8 close parenthesis close parenthesis$
$log sub 2 ⁡ open parenthesis 4 close parenthesis plus log sub 2 ⁡ open parenthesis 8 close parenthesis$

(a) and (c) are equal.

$log sub 2 ⁡ open parenthesis 16 plus 16 close parenthesis$
$log sub 2 ⁡ open parenthesis 16 close parenthesis plus log sub 2 ⁡ open parenthesis 16 close parenthesis$
$log sub 2 ⁡ open parenthesis 2 close parenthesis plus log sub 2 ⁡ open parenthesis 16 close parenthesis$

$log sub 3 ⁡ open parenthesis 27 squared close parenthesis$
$open parenthesis log sub 3 ⁡ open parenthesis 27 close parenthesis close parenthesis squared$
$log sub 3 ⁡ open parenthesis 27 close parenthesis plus log sub 3 ⁡ open parenthesis 27 close parenthesis$

(a) and (c) are equal.

$log sub 3 ⁡ open parenthesis 3 times 27 close parenthesis$
$log sub 3 ⁡ open parenthesis 3 close parenthesis plus log sub 3 ⁡ open parenthesis 27 close parenthesis$
$log sub 3 ⁡ open parenthesis 3 close parenthesis times log sub 3 ⁡ open parenthesis 27 close parenthesis$

$\log_{10} (\frac{240}{10})$
$\frac{\log_{10} (240)}{\log_{10} (10)}$
$log sub 10 ⁡ open parenthesis 240 minus 10 close parenthesis$

$log ⁡ 24 approximately equals 1.38$
$log ⁡ 240 approximately equals 2.38$
$log ⁡ 230 approximately equals 2.36$

None are equal.

$\log_{10} (\frac{1}{2} \cdot 80)$
$\frac{1}{2} \log_{10} (80)$
$log sub 10 ⁡ open parenthesis the square root of 80 close parenthesis$

$log sub 10 ⁡ open parenthesis 75 minus 15 close parenthesis$
$log sub 10 ⁡ open parenthesis 75 close parenthesis minus log sub 10 ⁡ open parenthesis 15 close parenthesis$
$\frac{\log_{10} (75)}{\log_{10} (15)}$

$log ⁡ open parenthesis 60 close parenthesis approximately equals 1.78$
$log ⁡ open parenthesis 5 close parenthesis approximately equals 0.70$
$\frac{\log_{10} (75)}{\log_{10} (15)} \approx 1.59$

None are equal.

$log sub 10 ⁡ open parenthesis 8 times 25 close parenthesis$
$log sub 10 ⁡ open parenthesis 25 to the power 8 close parenthesis$
$log sub 10 ⁡ open parenthesis 8 close parenthesis plus 25$

For Problems 53–58, use the formula for compound interest,

$A = P {(1 + \frac{r}{n})}^{n t}$

What rate of interest is required so that $$ 1000$ will yield $$ 1900$ after $5$ years if the interest rate is compounded monthly?

$12.9 %$

What rate of interest is required so that $$ 400$ will yield $$ 600$ after $3$ years if the interest rate is compounded quarterly?

How long will it take a sum of money to triple if it is invested at $10 %$ compounded daily?

About $11$ years

How long will it take a sum of money to increase by a factor of $5$ if it is invested at $10 %$ compounded quarterly?

Suppose you invest $$ 1000$ at $12 %$ annual interest for $5$ years. In this problem, we will investigate how the number of compounding periods, $n$ , affects the amount, $A$ . Write $A$ as a function of $n$ , with $P equals 1000$ , $r equals 0.12$ , and $t equals 5$ .
Use your calculator to make a table of values for $A$ as a function of $n$ . What happens to $A$ as $n$ increases?
What value of $n$ is necessary to produce an amount $A greater than 1818$ ? To produce $A greater than 1820$ ? To produce $A greater than 1822$ ?
Graph the function $A (n)$ in the window
$\begin{aligned} Xmin & = 0 & Xmax = 52 \\ Ymin & = 1750 & Ymax = 1850 \end{aligned}$
Describe the graph: Is it increasing or decreasing? Concave up or down? Does it appear to have an asymptote? Give your best estimate for the asymptote.

$A = 1000 {(1 + \frac{0.12}{n})}^{5 n}$
$A$ increases.
$16$ ; $31$ ; $553$
Increasing, concave down, asymptotically approaching $A approximately equals 1822.12$

In this problem we will repeat Problem 57 for $4 %$ interest. Write $A$ as a function of $n$ , with $P equals 1000$ , $r equals 0.04$ , and $t equals 5$ .
Use your calculator to make a table of values for $A$ as a function of $n$ . What happens to $A$ as $n$ increases?
What value of $n$ is necessary to produce an amount $A greater than 1218$ ? To produce $A greater than 1220$ ? To produce $A greater than 1221.40$ ?
Graph the function $A (n)$ in the window
$\begin{aligned} Xmin & = 0 & Xmax = 52 \\ Ymin & = 1210 & Ymax = 1225 \end{aligned}$
Describe the graph: Is it increasing or decreasing? Concave up or down? Does it appear to have an asymptote? Give your best estimate for the asymptote.

For Problems 59–64, solve the formula for the specified variable.

$N = N_{0} a^{k t}$ , for $k$

$k = \frac{1}{t} \frac{\log (N / N_{0})}{\log (a)}$

$Q = Q_{0} b^{t / 2}$ , for $t$

$A = A_{0} (10^{k t} - 1)$ , for $t$

$t = \frac{1}{k} \log (\frac{A}{A_{0}} + 1)$

$B = B_{0} (1 - 10^{- k t})$ , for $t$

$w = p v^{q}$ , for $q$

$q = \frac{\log (w / p)}{\log (v)}$

$l = p^{a} q^{b}$ , for $b$

In Problems 65–68 we use the laws of exponents to prove the properties of logarithms.

We will use the first law of exponents, $a^{p} \cdot a^{q} = a^{p + q}$ , to prove the first property of logarithms.

Let $m = \log_{b} (x)$ and $n = \log_{b} (y)$ . Rewrite these equations in exponential form:
$x = \underset{―}{} and y = \underset{―}{}$
Now consider the expression $\log_{b} (x y)$ . Replace $x$ and $y$ by your answers to part (a).
Apply the first law of exponents to your expression in part (b).
Use the definition of logarithm to simplify your answer to part (c).
Refer to the definitions of $m$ and $n$ in part (a) to finish the proof.

$x = b^{m}$ , $y = b^{n}$
$\log_{b} (b^{m} \cdot b^{n})$
$\log_{b} (b^{m} \cdot b^{n}) = \log_{b} (b^{m + n})$
$\log_{b} (b^{m + n}) = m + n$
$\log_{b} (b^{m + n}) = (\log_{b} (x)) + (\log_{b} (y))$

We will use the second law of exponents, $\frac{a^{p}}{a^{q}} = a^{p - q}$ , to prove the second property of logarithms.

Let $m = \log_{b} (x)$ and $n = \log_{b} (y)$ . Rewrite these equations in exponential form:
$x = \underset{―}{} and y = \underset{―}{}$
Now consider the expression $\log_{b} (\frac{x}{y})$ . Replace $x$ and $y$ by your answers to part (a).
Apply the second law of exponents to your expression in part (b).
Use the definition of logarithm to simplify your answer to part (c).
Refer to the definitions of $m$ and $n$ in part (a) to finish the proof.

We will use the third law of exponents, $(a^{p})^{q} = a^{p q}$ , to prove the third property of logarithms.

Let $m = \log_{b} (x)$ . Rewrite this equation in exponential form:
$x = \underset{―}{}$
Now consider the expression $\log_{b} (x^{k})$ . Replace $x$ by your answers to part (a).
Apply the third law of exponents to your expression in part (b).
Use the definition of logarithm to simplify your answer to part (c).
Refer to the definitions of $m$ in part (a) to finish the proof.

$x = b^{m}$
$\log_{b} ((b^{m})^{k})$
$\log_{b} ((b^{m})^{k}) = \log_{b} (b^{m k})$
$\log_{b} (b^{m k}) = m k$
$\log_{b} (b^{m k}) = (\log_{b} (x)) \cdot k$

Use the logarithms and exponents conversion equations to explain why $log sub b ⁡ open parenthesis 1 close parenthesis equals 0$ .
Use the logarithms and exponents conversion equations to explain why $\log_{b} (b^{x}) = x$ .
Use the logarithms and exponents conversion equations to explain why $b^{\log_{b} (x)} = x$ .

Modeling, Functions, and Graphs by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.