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Section 4.3 Logarithms

Subsection Introduction

In this section, we introduce a new mathematical tool called a logarithm, which will help us solve exponential equations.

Suppose that a colony of bacteria doubles in size every day. If the colony starts with \(50\) bacteria, how long will it be before there are \(800\) bacteria? We answered questions of this type in Section 4.2 by writing and solving an exponential equation. The function

\begin{equation*} P(t) = 50 \cdot 2^t \end{equation*}

gives the number of bacteria present on day \(t\text{,}\) so we must solve the equation

\begin{equation*} 800 = 50 \cdot 2^t \end{equation*}

Dividing both sides by 50 yields

\begin{equation*} 16 = 2^t \end{equation*}

The solution of this equation is the answer to the following question:

To what power must we raise \(2\) in order to get \(16\text{?}\)

The value of \(t\) that solves the equation is called the base \(2\) logarithm of \(16\text{.}\) Because \(2^4 = 16\text{,}\) the base \(2\) logarithm of \(16\) is \(4\text{.}\) We write this as

\begin{equation*} \log_{2}16 = 4 \end{equation*}

In other words, we solve an exponential equation by computing a logarithm. You can check that \(t = \alert{4}\) solves the problem stated above:

\begin{equation*} P(\alert{4}) = 50 \cdot 2^{\alert{4}}= 800 \end{equation*}

Thus, the unknown exponent is called a logarithm. In general, for positive values of \(b\) and \(x\text{,}\) we make the following definition.

Definition of Logarithm

For \(b\gt 0, b\ne 1\text{,}\) the base \(b\) logarithm of \(x\), written \(\log_{b} x\text{,}\) is the exponent to which \(b\) must be raised in order to yield \(x\text{.}\)

Note 4.29

It will help to keep in mind that a logarithm is just an exponent.

Some logarithms, like some square roots, are easy to evaluate, while others require a calculator. We will start with the easy ones.

Example 4.30

Compute the logarithms.

  1. \(\log_3 9\)

  2. \(\log_5 125 \)

  3. \(\log_4 \dfrac{1}{16}\)

  4. \(\log_5 \sqrt{5}\)

Solution
  1. To evaluate \(\log_3 9\text{,}\) we ask what exponent on base \(3\) will produce \(9\text{.}\) Or, 3 to what power equals 9? The exponent we need is \(\alert{2}\text{,}\) so

    \begin{equation*} \log_3 9 = \alert{2}~~ \text{ because } ~~3^\alert{2} = 9 \end{equation*}

    We use similar reasoning to compute the other logarithms.

  2. \(\log_5{125} = \alert{3}~~ \text{ because } ~~5^\alert{3} = 125\)

  3. \(\log_4{\dfrac{1}{16}}= \alert{-2} ~~\text{ because } ~~4^{\alert{-2}} = \dfrac{1}{16}\)

  4. \(\log_5{\sqrt{5}} = \alert{\dfrac{1}{2}}~~ \text{ because }~~ 5^{\alert{1/2}} =\sqrt{5}\)

Checkpoint 4.31

Find each logarithm.

  1. \(\log_{3}{81}\)

  2. \(\log_{10}{\dfrac{1}{1000}}\)

Answer
  1. \(4\)

  2. \(-3\)

From the definition of a logarithm and the examples above, we see that the following two statements are equivalent.

Logarithms and Exponents: Conversion Equations

If \(b \gt 0\text{,}\) \(b\ne 1\text{,}\) and \(x \gt 0\text{,}\)

\begin{equation*} \blert{y = \log_b x}~~~ \text{ if and only if }~~~ \blert{ x = b^y} \end{equation*}

In other words, the logarithm, \(y\text{,}\) is the same as the exponent in \(x = b^y\text{.}\) We see again that a logarithm is an exponent; it is the exponent to which \(b\) must be raised to yield \(x\text{.}\)

These equations allow us to convert from logarithmic to exponential form, or vice versa. You should memorize the conversion equations, because we will use them frequently.

As special cases of the equivalence in (1), we can compute the following useful logarithms. For any base \(b \gt 0, b\ne 1\text{,}\)

Some Useful Logarithms
\begin{equation*} \begin{aligned}[t] \log_b b \amp = 1~~~ \text{ because } ~~~b^1 = b \\ \log_b 1 \amp = 0 ~~~ \text{ because } ~~~b^0 = 1 \\ \log_b{b^x} \amp = x~~~ \text{ because } ~~~b^x = b^x \end{aligned} \end{equation*}
Example 4.32
  1. \(\log_{2}{2} = 1\)

  2. \(\log_{5}{1} = 0\)

  3. \(\log_{3}{3^4} = 4\)

Checkpoint 4.33

Find each logarithm.

  1. \(\log_{n}{1}\)

  2. \(\log_{n}{n^3}\)

Answer
  1. \(0\)

  2. \(3\)

Subsection Using the Conversion Equations

We use logarithms to solve exponential equations, just as we use square roots to solve quadratic equations. Consider the two equations

\begin{equation*} x^2 = 25 ~~~~ \text{ and } ~~~~ 2^x = 8 \end{equation*}

We solve the first equation by taking a square root, and we solve the second equation by computing a logarithm:

\begin{equation*} x = \pm\sqrt{25} = \pm 5 ~~~~ \text{ and } ~~~~ x = \log_{2}{8} = 3 \end{equation*}

The operation of taking a base \(b\) logarithm is the inverse operation for raising the base \(b\) to a power, just as extracting square roots is the inverse of squaring a number.

Every exponential equation can be rewritten in logarithmic form by using the conversion equations. Thus,

\begin{equation*} 3 = \log_{2}{8}~~~~ \text{ and }~~~~ 8 = 2^3 \end{equation*}

are equivalent statements, just as

\begin{equation*} 5 = \sqrt{25}~~~~ \text{ and }~~~~ 25 = 5^2 \end{equation*}

are equivalent statements. Rewriting an equation in logarithmic form is a basic strategy for finding its solution.

Example 4.34

Rewrite each equation in logarithmic form.

  1. \(2^{-1} = \dfrac{1}{2}\)

  2. \(a^{1/5} = 2.8\)

  3. \(6^{1.5} = T\)

  4. \(M^v = 3K\)

Solution

First identify the base \(b\text{,}\) and then the exponent or logarithm \(y\text{.}\) Use the conversion equations to rewrite \(b^y = x\) in the form \(\log_{b}{x} = y\text{.}\)

  1. The base is \(2\) and the exponent is \(-1\text{.}\) Thus, \(\log_{2}{\dfrac{1}{2}}= -1\text{.}\)

  2. The base is \(a\) and the exponent is \(\dfrac{1}{5}\text{.}\) Thus, \(\log_{a}{2.8} = \dfrac{1}{5}\text{.}\)

  3. The base is \(6\) and the exponent is \(1.5\text{.}\) Thus, \(\log_{6}{T} = 1.5\text{.}\)

  4. The base is \(M\) and the exponent is \(v\text{.}\) Thus, \(\log_{M}{3K} = v\text{.}\)

Checkpoint 4.35

Rewrite each equation in logarithmic form.

  1. \(8^{-1/3} = \dfrac{1}{2}\)

  2. \(5^x = 46\)

Answer
  1. \(\log_8 \left(\dfrac{1}{2} \right)=\dfrac{-1}{3} \)

  2. \(\log_{5} 46=x \)

Subsection Approximating Logarithms

Suppose we would like to solve the equation

\begin{equation*} 2^x = 26 \end{equation*}

The solution of this equation is \(x = \log_{2}{26}\text{,}\) but can we find a decimal approximation for this value? There is no integer power of \(2\) that equals \(26\text{,}\) because

\begin{equation*} \begin{aligned}[t] 2^4 \amp = 16 \\ \text{and }~~~~ 2^5 \amp = 32 \end{aligned} \end{equation*}

Thus, \(\log_{2}{26}\) must be between \(4\) and \(5\text{.}\) We can use trial and error to find the value of \(\log_{2}{26}\) to the nearest tenth. Use your calculator to make a table of values for \(y = 2^x\text{,}\) starting with \(x = 4\) and using increments of \(0.1\text{.}\)

\(x\) \(2^x\) \(x\) \(2^x\)
\(4\) \(2^4=16\) \(4.5\) \(2^{4.5}=22.627\)
\(4.1\) \(2^{4.1}=17.148\) \(4.6\) \(2^{4.6}=24.251\)
\(4.2\) \(2^{4.2}=18.379\) \(\alert{4.7}\) \(2^{\alert{4.7}}=25.992\)
\(4.3\) \(2^{4.3}=19.698\) \(\alert{4.8}\) \(2^{\alert{4.8}}=27.858\)
\(4.4\) \(2^{4.4}=21.112\) \(4.9\) \(2^{4.9}=29.857\)

From the table we see that \(26\) is between \(2^{4.7}\) and \(2^{4.8}\text{,}\) and is closer to \(2^{4.7}\text{.}\) To the nearest tenth, \(\log_{2}{26} \approx 4.7\text{.}\)

Trial and error can be a time-consuming process. In Example 4, we illustrate a graphical method for estimating the value of a logarithm.

Example 4.36

Approximate \(\log_{3}{7}\) to the nearest hundredth.

Solution

If \(\log_{3}{7}=x\text{,}\) then \(3^x = 7\text{.}\) We will use the graph of \(y = 3^x\) to approximate a solution to \(3^x = 7\text{.}\)

We graph \(Y_1 =3\)^ X and \(Y_2 = 7\) in the standard window (ZOOM 6) to obtain the graph shown below. Next we activate the intersect feature to find that the two graphs intersect at the point \((1.7712437, 7)\text{.}\) Because this point lies on the graph of \(y = 3^x\) , we know that

\begin{equation*} 3^{1.7712437} \approx 7~~~\text{, or }~~~ \log_{3}{7} \approx 1.7712437 \end{equation*}

To the nearest hundredth, \(\log_{3}{7} \approx 1.77\text{.}\)

GC intersection of expnential curve and horizontal line
Checkpoint 4.37
  1. Rewrite the equation \(3^x = 90\) in logarithmic form.

  2. Use a graph to approximate the solution to the equation in part (a). Round your answer to three decimal places.

Answer
  1. \(\log_8(90)=x \)

  2. \(x\approx 4.096 \)

Subsection Base 10 Logarithms

Some logarithms are used so frequently in applications that their values are programmed into scientific and graphing calculators. These are the base \(10\) logarithms, such as

\begin{equation*} \log_{10}{1000} = 3 ~~~\text{ and }~~~ \log_{10}{0.01} = -2 \end{equation*}

Base \(10\) logarithms are called common logarithms, and the subscript \(10\) is often omitted, so that \(\log x\) is understood to mean \(\log_{10}{x}\text{.}\)

To evaluate a base \(10\) logarithm, we use the LOG key on a calculator. Many logarithms are irrational numbers, and the calculator gives as many digits as its display allows. We can then round off to the desired accuracy.

Example 4.38

Approximate the following logarithms to \(2\) decimal places.

  1. \(\log{6.5}\)

  2. \(\log{256}\)

Solution
  1. The keying sequence LOG \(6.5\) )ENTER produces the display

    \(\log {(6.5)}\) \(\) \(\)
    \(\) \(\) \(.812913566\)

    so \(\log {6.5}\approx 0.81\text{.}\)

  2. The keying sequence LOG \(256\) ) ENTER yields \(2.408239965\text{,}\) so \(\log {256} \approx 2.41\text{.}\)

Note 4.39

We can check the approximations found in Example 4.38 with our conversion equations. Remember that a logarithm is an exponent, and in this example the base is \(10\text{.}\) We find that

\begin{equation*} \begin{aligned}[t] \amp\amp 10^{0.81} \amp\approx 6.45654229 \\ \text{and} \amp\amp 10^{2.41} \amp\approx 257.0395783 \end{aligned} \end{equation*}

so our approximations are reasonable, although you can see that rounding a logarithm to \(2\) decimal places does lose some accuracy.

For this reason, rounding logarithms to \(4\) decimal places is customary.

Checkpoint 4.40
  1. Evaluate \(\log{250}\text{,}\) and round your answer to two decimal places. Check your answer using the conversion equations.

  2. Evaluate \(\log{250}\text{,}\) and round your answer to four decimal places. Check your answer using the conversion equations.

Answer
  1. \(2.40\)

  2. \(2.3979\)

Subsection Solving Exponential Equations

We can now solve any exponential equation with base \(10\text{.}\) For instance, to solve the equation \(16 \cdot 10^t = 360\text{,}\) we first divide both sides by \(16\) to obtain

\begin{equation*} 10^t = 22.5 \end{equation*}

Then we convert the equation to logarithmic form and evaluate:

\begin{equation*} t = \log_{10}{22.5} \approx 1.352182518 \end{equation*}

To \(4\) decimal places, the solution is \(1.3522\text{.}\)

To solve exponential equations involving powers of 10, we can use the following steps.

Steps for Solving Base 10 Exponential Equations
  1. Isolate the power on one side of the equation.

  2. Rewrite the equation in logarithmic form.

  3. Use a calculator, if necessary, to evaluate the logarithm.

  4. Solve for the variable.

Example 4.41

Solve the equation \(~~38 = 95 - 15 \cdot 10^{0.4x}\)

Solution

First, we isolate the power of \(10\text{:}\) We subtract \(95\) from both sides of the equation and divide by \(-15\) to obtain

\begin{equation*} \begin{aligned}[t] -57 \amp = -15 \cdot 10^{0.4x} \amp\amp \blert{\text{Divide by }-15.} \\ 3.8 \amp = 10^{0.4x} \end{aligned} \end{equation*}

Next, we convert the equation to logarithmic form as

\begin{equation*} \log_{10}{3.8} = 0.4x \end{equation*}

Solving for \(x\) yields

\begin{equation*} \frac{\log_{10}{3.8}}{0.4}= x \end{equation*}

We can evaluate this expression on the calculator by entering

LOG \(3.8\) ) ÷ \(0.4\) ENTER

which yields \(1.449458992\text{.}\) Thus, to four decimal places, \(x \approx 1.4495\text{.}\)

Caution 4.42

Be careful when using a calculator to evaluate expressions involving logs. We can evaluate a single logarithm like \(\log{3.8}\) by entering LOG \(3.8\)ENTER without an ending parenthesis, so that the calculator shows

\(\log {(3.8}\) \(\) \(\)
\(\) \(\) \(.5795835966\)

But if we want to evaluate \(\dfrac{\log{3.8}}{0.4}\text{,}\) we must enclose \(3.8\) in parentheses, as shown in Example 4.41. If we omit the parenthesis after \(3.8\text{,}\) the calculator will interpret the expression as \(\log{\left(\dfrac{3.8}{0.4}\right)}\text{,}\) which is not the expression we wanted.

Checkpoint 4.43

Solve \(~~12 - 30(10^{-0.2x} ) = 11.25\)

Answer

\(8.01\)

Subsection Application to Exponential Models

We have seen that exponential functions are used to describe some applications of growth and decay, \(P(t) = P_0b^t\text{.}\) There are two common questions that arise in connection with exponential models:

  1. Given a value of \(t\text{,}\) what is the corresponding value of \(P(t)\text{?}\)

  2. Given a value of \(P(t)\text{,}\) what is find the corresponding value of \(t\text{?}\)

To answer the first question, we evaluate the function \(P(t)\) at the appropriate value. To answer the second question, we must solve an exponential equation, and this usually involves logarithms.

Example 4.44

The value of a large tractor originally worth $\(30,000\) depreciates exponentially according to the formula

\begin{equation*} V(t) = 30,000(10)^{-0.04t} \end{equation*}

where \(t\) is in years. When will the tractor be worth half its original value?

Solution

We want to find the value of \(t\) for which \(V(t) = 15,000\text{.}\) That is, we want to solve the equation

\begin{equation*} 15,000 = 30,000(10)^{-0.04t} \end{equation*}

We divide both sides by 30,000 to obtain

\begin{equation*} 0.5 = 10^{-0.04t} \end{equation*}

We convert the equation to logarithmic form as

\begin{equation*} \log_{10}{0.5} = -0.04t \end{equation*}

and divide by \(-0.04\) to obtain

\begin{equation*} \frac{\log_{10}{0.5}}{-0.04} = t \end{equation*}

To evaluate this expression, we key in

LOG \(0.5\) ) ÷ (-) \(0.04\) ENTER

to find \(t \approx 7.525749892\text{.}\) The tractor will be worth $15,000 in approximately \(7\frac{1}{2}\) years.

Checkpoint 4.45

The percentage of American homes with computers grew exponentially from 1994 to 1999. For \(t = 0\) in 1994, the growth law was

\begin{equation*} P(t) = 25.85(10)^{0.052t} \end{equation*}

[Source: Los Angeles Times, August 20, 1999]

  1. What percent of American homes had computers in 1994?

  2. If the percentage of homes with computers continued to grow at the same rate, when did \(90\)% of American homes have a computer?

  3. Do you think that the function \(P(t)\) will continue to model the percentage of American homes with computers? Why or why not?

Answer
  1. \(25.85\%\)

  2. \(t\approx 10.4\) (year 2004)

  3. No, the percent of homes with computers cannot exceed \(100\%\text{.}\)

At this stage, it seems we will only be able to solve exponential equations in which the base is \(10\text{.}\) However, we will see in Section 4.4 how the properties of logarithms enable us to solve exponential equations with any base.

Subsection Section Summary

Subsubsection Vocabulary

Look up the definitions of new terms in the Glossary.

  • Logarithm

  • Common logarithm

Subsubsection CONCEPTS

  1. We use logarithms to help us solve exponential equations.

  2. The base \(b\) logarithm of \(x\), written \(\log_b x\text{,}\) is the exponent to which \(b\) must be raised in order to yield \(x\text{.}\)

  3. If \(b\gt 0\) and \(x \gt 0\text{,}\)

    \begin{equation*} y=\log_b x ~~~ \text{ if and only if } ~~~x=b^y \end{equation*}
  4. The graphs of exponential functions can be transformed by shifts, stretches, and reflections.

  5. The operation of taking a base \(b\) logarithm is the inverse operation for raising the base \(b\) to a power.

  6. Base \(10\) logarithms are called common logarithms, and \(\log x\) means \(\log_{10} x\text{.}\)

  7. Steps for Solving Base 10 Exponential Equations
    1. Isolate the power on one side of the equation.

    2. Rewrite the equation in logarithmic form.

    3. Use a calculator, if necessary, to evaluate the logarithm.

    4. Solve for the variable.

Subsubsection STUDY QUESTIONS

  1. To find \(\log_6 27\) means to find an exponent \(\) that satisfies the equation .

  2. Can a logarithm be a negative number?

  3. Evaluate the following logarithms:

    1. \(\log_8 8^{15}\)

    2. \(\log_5 5^{\sqrt{13}}\)

    3. \(\log_b b^{2.63}\)

  4. Guess the solution of \(10^x = 750\text{.}\) Now find an approximation correct to four decimal places. Was your guess too big or too small?

Subsubsection SKILLS

Practice each skill in the Homework  problems listed.

  1. Compute logs base \(b\) using the definition: #1–10, 59–66

  2. Convert from exponential to logarithmic form: #11–22

  3. Approximate logarithms: #23–34

  4. Solve exponential equations base 10: #35–48

  5. Solve application problems: #49–58

Subsection Homework 4.3

For Problems 1-10, find each logarithm without using a calculator.

1
  1. \(\log_7 49\)

  2. \(\log_2 32 \)

2
  1. \(\log_4 64\)

  2. \(\log_3 27 \)

3
  1. \(\log_3 \sqrt{3} \)

  2. \(\log_3 \dfrac{1}{3} \)

4
  1. \(\log_5 \dfrac{1}{5} \)

  2. \(\log_5 \sqrt{5} \)

5
  1. \(\log_4 4 \)

  2. \(\log_6 1 \)

6
  1. \(\log_{10} 1 \)

  2. \(\log_{10} 10^{-6} \)

7
  1. \(\log_8 8^5 \)

  2. \(\log_7 7^6 \)

8
  1. \(\log_{10} 10^{-4} \)

  2. \(\log_{10} 10^{-6} \)

9
  1. \(\log_{10} 0.1 \)

  2. \(\log_{10} 0.001 \)

10
  1. \(\log_{10} 10,000 \)

  2. \(\log_{10} 1000 \)

For Problems 11-22, rewrite the equation in logarithmic form.

11

\(2^{10}=1024 \)

12

\(11^{4}=14,641 \)

13

\(10^{0.699}\approx 5 \)

14

\(10^{-0.602}\approx 0.25 \)

15

\(t^{3/2}=16 \)

16

\(v^{5/3}=12 \)

17

\(0.8^{1.2}=M \)

18

\(3.7^{2.5}=Q \)

19

\(x^{5t}=W-3 \)

20

\(z^{-3t}=2P+5 \)

21

\(3^{-0.2t}=2N_0 \)

22

\(10^{1.3t}=3M_0 \)

For Problems 23–26,

  1. Solve each equation, writing your answer as a logarithm.

  2. Use trial and error to approximate the logarithm to one decimal place.

23

\(4^x=2.5 \)

24

\(2^x=0.2 \)

25

\(10^x=0.003 \)

26

\(10^x=4500 \)

For Problems 27–30,

  1. By computing successive powers of the base, trap each log between two integers.

  2. Use a graph to approximate each logarithm to the nearest hundredth. (Hint: Use the conversion equations to rewrite \(x = \log_b y\) as an appropriate exponential equation.)

27

\(\log_{10} 7 \)

28

\(\log_{10} 50 \)

29

\(\log_{3} 67.9 \)

30

\(\log_{5} 86.3 \)

For Problems 31-34, use a calculator to approximate each logarithm to four decimal places. Make a conjecture about logarithms based on the results of each problem.

31
  1. \(\log_{10} 5.43\)

  2. \(\log_{10} 54.3\)

  3. \(\log_{10} 543\)

  4. \(\log_{10} 5430\)

32
  1. \(\log_{10} 0.625\)

  2. \(\log_{10} 0.0625\)

  3. \(\log_{10} 0.00625\)

  4. \(\log_{10} 0.000625\)

33
  1. \(\log 2\)

  2. \(\log 4 \)

  3. \(\log 8 \)

  4. \(\log 16\)

34
  1. \(\log 4\)

  2. \(\log 0.25\)

  3. \(\log 5\)

  4. \(\log 0.2\)

For Problems 35-44, solve for \(x\text{.}\) Round your answers to hundredths.

35

\(10^{-3x}=5 \)

36

\(10^{-5x}=76 \)

37

\(25\cdot 10^{0.2x}=80 \)

36

\(8\cdot 10^{1.6x}=312 \)

39

\(12.2=2\left(10^{1.4x}\right)-11.6 \)

40

\(163=3\left(10^{0.7x}\right)-49.3 \)

41

\(3\left(10^{-1.5x}\right)-14.7=17.1 \)

42

\(4\left(10^{-0.6x}\right)+16.1=28.2 \)

43

\(80\left(1-10^{-0.2x}\right) =65 \)

44

\(250\left(1-10^{-0.3x}\right) =100 \)

In Problems 45–48, each calculation contains an error. Find and correct it.

45
\begin{equation*} \begin{aligned}[t] 2\cdot 5^x\amp =848 \amp\amp\\ 10^x\amp = 848\\ x\amp=\log 848 \amp\amp\bluetext{(Incorrect!)} \end{aligned} \end{equation*}
46
\begin{equation*} \begin{aligned}[t] 15\cdot 10^x\amp =20 \amp\amp\\ 10^x\amp = 5\\ x\amp=\log 5 \amp\amp\bluetext{(Incorrect!)} \end{aligned} \end{equation*}
47
\begin{equation*} \begin{aligned}[t] 10^{4x}\amp =20 \amp\amp\\ 10^x\amp = 5\\ x\amp=\log 5 \amp\amp\bluetext{(Incorrect!)} \end{aligned} \end{equation*}
48
\begin{equation*} \begin{aligned}[t] 12+ 6^x\amp =42 \amp\amp\\ 6^x\amp = 30\\ x\amp= 5 \amp\amp\bluetext{(Incorrect!)} \end{aligned} \end{equation*}
49

The population of the state of California increased during the years 1990 to 2000 according to the formula

\begin{equation*} P(t) = 29,760,021 (10)^{0.0056t}\text{,} \end{equation*}

where \(t\) is measured in years since \(1990\text{.}\)

  1. What was the population in \(2000\text{?}\)

  2. Assuming the same rate of growth, estimate the population of California in the years \(2010\text{,}\) \(2015\text{,}\) and \(2020\text{.}\)

  3. When did the population of California reach \(35,000,000\text{?}\)

  4. When should the population reach \(40,000,000\text{?}\)

  5. Graph the function \(P\) with a suitable domain and range, then verify your answers to parts (a) through (d).

50

The population of the state of New York increased during the years 1990 to 2000 according to the formula

\begin{equation*} P(t) = 17,990,455 (10)^{0.0023t}\text{,} \end{equation*}

where \(t\) is measured in years since \(1990\text{.}\)

  1. What was the population in \(2000\text{?}\)

  2. Assuming the same rate of growth, estimate the population of New York in the years \(2010\text{,}\) \(2015\text{,}\) and \(2020\text{.}\)

  3. When did the population of New York reach \(20,000,000\text{?}\)

  4. When should the population reach \(30,000,000\text{?}\)

  5. Graph the function \(P\) with a suitable domain and range, then verify your answers to parts (a) through (d).

51

The absolute magnitude, \(M\text{,}\) of a star is a measurement of its brightness. For example, our Sun, not a particularly bright star, has magnitude \(M = 4.83\text{.}\) The magnitude in turn is a measure of the luminosity, \(L\text{,}\) or amount of light energy emitted by the star, where

\begin{equation*} L = L_0 10^{-0.4M} \end{equation*}
  1. The luminosity of a star is measured in solar units, so that our Sun has luminosity \(L = 1\text{.}\) Use the values of \(L\) and \(M\) for the Sun to calculate a value of \(L_0\) in the equation above.

  2. Is luminosity an increasing or decreasing function of magnitude? Graph the function on the domain \([-3, 3]\text{.}\) What is its range on that domain?

  3. The luminosity of Sirius is \(22.5\) times that of the Sun, or \(L = 22.5\text{.}\) Calculate the magnitude of Sirius.

  4. If two stars differ in magnitude by \(5\text{,}\) what is the ratio of their luminosities?

  5. A decrease in magnitude by \(1\) corresponds to an increase in luminosity by what factor? Give an exact value and an approximation to four decimal places.

  6. Normal stars have magnitudes between \(-10\) and \(19\text{.}\) What range of luminosities do stars exhibit?

52

The loudness of a sound is a consequence of its intensity, \(I\text{,}\) or the amount of energy it generates, in watts per square meter. The intensity is related to the decibel level, \(D\text{,}\) which is another measure of loudness, by

\begin{equation*} I = 10^{-12+D/10} \end{equation*}
  1. Is intensity an increasing or decreasing function of decibel level? The faintest sound a healthy human can hear is \(0\) decibels. What is the intensity of a \(0\) decibel sound?

  2. A whisper produces an energy intensity of \(10^{-9}\) watts per square meter. What is the decibel level of a whisper?

  3. If two sounds differ in loudness by \(10\) decibels, what is the ratio of their intensities?

  4. An increase in loudness of \(1\) decibel produces a just noticeable difference to the human ear. By what factor does the intensity increase?

  5. Sounds of \(130\) decibels are at the threshold of pain for people. What is the range of the intensity function on the domain \([0, 130]\text{?}\)

The atmospheric pressure decreases with altitude above the surface of the Earth. For Problems 53–58, use the function

\begin{equation*} P(h) = 30 (10)^{-0.09h} \end{equation*}

where altitude, \(h\text{,}\) is given in miles and atmospheric pressure, \(P\text{,}\) in inches of mercury. Graph this function in the window

\begin{align*} {\text{Xmin}} \amp = 0 \amp\amp {\text{Xmax}} = 9.4\\ {\text{Ymin}} \amp = 0 \amp\amp {\text{Ymax}} = 30 \end{align*}

Solve the problems below algebraically, and verify with your graph.

53

The elevation of Mount Everest, the highest mountain in the world, is \(29,028\) feet. What is the atmospheric pressure at the top?

Hint

\(1\) mile \(= 5280\) feet

54

The elevation of Mount McKinley, the highest mountain in the United States, is \(20,320\) feet. What is the atmospheric pressure at the top?

55

How high above sea level is the atmospheric pressure \(20.2\) inches of mercury?

56

How high above sea level is the atmospheric pressure \(16.1\) inches of mercury?

57

Find the height above sea level at which the atmospheric pressure is equal to one-half the pressure at sea level.

Hint

What is the altitude at sea level?

58

Find the height above sea level at which the atmospheric pressure is equal to one-fourth the pressure at sea level.

Hint

What is the altitude at sea level?

For Problems 59-66, simplify the expression.

59

\(\log_2 (\log_4 16) \)

60

\(\log_5 (\log_5 5) \)

61

\(\log_{10} [\log_3(\log_5 125) ] \)

62

\(\log_{10} [\log_2 (\log_3 9)] \)

63

\(\log_2[\log_2 (\log_3 81)] \)

64

\(\log_4 [\log_2 (\log_3 81)] \)

65

\(\log_{b} (\log_b b) \)

66

\(\log_{b} (\log_a a^b) \)