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Modeling, Functions, and Graphs

Katherine Yoshiwara

Section 3.4 Rational Exponents

Subsection Powers of the Form \(a^{m/n}\)

In the last section, we considered powers of the form \(a^{1/n}\text{,}\) such as \(x^{1/3}\) and \(x^{-1/4}\text{,}\) and saw that \(a^{1/n}\) is equivalent to the root \(\sqrt[n]{a}\text{.}\) What about other fractional exponents? What meaning can we attach to a power of the form \(a^{m /n}\text{?}\)
Consider the power \(x^{3/2}\text{.}\) Notice that the exponent \(\dfrac{3}{2}= 3(\dfrac{1}{2})\text{,}\) and thus by the third law of exponents, we can write
\begin{equation*} (x^{1/2})^3 = x^{(1/2)^3} = x^{3/2} \end{equation*}
In other words, we can compute \(x^{3/2}\) by first taking the square root of \(x\) and then cubing the result. For example,
\begin{equation*} \begin{aligned}[t] 100^{3/2} \amp = (\alert{100^{1/2}})^3 \amp\amp \blert{\text{Take the square root of 100.}}\\ \amp= \alert{10^3} = 1000 \amp\amp \blert{\text{Cube the result.}} \end{aligned} \end{equation*}
We will define fractional powers only when the base is a positive number.

Rational Exponents.

\begin{equation*} a^{m/n} = (a^{1/n})^m = (a^m)^{1/n},~~~~a \gt 0\text,~~ n \ne 0 \end{equation*}

Checkpoint 3.82. QuickCheck 1.

\(x^{\frac{3}{4}}\) means
  • the fourth root of \(x\) cubed
  • the cube root of \(x\) to the fourth
  • the 3/4 root of \(x\)
  • the cube root of the fourth root of \(x\)
Answer.
\(\text{Choice 1}\)
Solution.
\(x^{\frac{3}{4}}\) means the fourth root of \(x\) cubed
To compute \(a^{m/n}\text{,}\) we can compute the \(n\)th root first, or the \(m\)th power, whichever is easier. For example,
\begin{equation*} 8^{2/3} = \left(8^2\right)^{1/3} = 64^{1/3} = 4 \end{equation*}
or
\begin{equation*} 8^{2/3} = \left(8^{1/3}\right)^2 = 2^2 = 4 \end{equation*}

Example 3.83.

  1. \begin{equation*} \begin{aligned}[t] 81^{3/4} \amp = \left(81^{1/4}\right)^3 \\ \amp = 3^3 = 27 \end{aligned} \end{equation*}
  2. \begin{equation*} \begin{aligned}[t] -27^{5/3} \amp = -\left(27^{1/3}\right)^5 \\ \amp = -3^5 = -243 \end{aligned} \end{equation*}
  3. \begin{equation*} \begin{aligned}[t] 27^{-2/3} \amp = \frac{1}{\left(27^{1/3}\right)^2} \\ \amp = \frac{1}{3^2}= \frac{1}{9} \end{aligned} \end{equation*}
  4. \begin{equation*} \begin{aligned}[t] 5^{3/2} \amp = \left(5^{1/2}\right)^3 \\ \amp \approx (2.236)^3 \approx 11.180 \end{aligned} \end{equation*}

Note 3.84.

You can verify all the calculations in Example 3.83 on your calculator. For example, to evaluate \(81^{3/4}\text{,}\) key in
81 ^ ( 3 ÷ 4 ) ENTER
or simply
81 ^ 0.75 ENTER

Checkpoint 3.85. Practice 1.

Evaluate each power.
  1. \(32^{-3/5}=\)
  2. \(-81^{1.25}=\)
Answer 1.
\({\frac{1}{8}}\)
Answer 2.
\(-243\)
Solution.
  1. \(\displaystyle \dfrac{1}{8}\)
  2. \(\displaystyle -243\)

Checkpoint 3.86. Pause and Reflect.

Explain how to evaluate \(f(x)=x^{\frac{-3}{4}}\) for \(x=625\) by hand.

Subsection Power Functions

The graphs of power functions \(y = x^{m/n}\text{,}\) where \(m/n\) is positive, are all increasing for \(x\ge 0\text{.}\) If \(m/n \gt 1\text{,}\) the graph is concave up. If \(0 \lt m/n \lt 1\text{,}\) the graph is concave down. Some examples are shown below.
power functions
Perhaps the single most useful piece of information a scientist can have about an animal is its metabolic rate. The metabolic rate is the amount of energy the animal uses per unit of time for its usual activities, including locomotion, growth, and reproduction. The basal metabolic rate, or BMR, sometimes called the resting metabolic rate, is the minimum amount of energy the animal can expend in order to survive.

Example 3.87.

A revised form of Kleiber’s rule states that the basal metabolic rate for many groups of animals is given by
\begin{equation*} B(m) = 70m^{0.75} \end{equation*}
where \(m\) is the mass of the animal in kilograms and the BMR is measured in kilocalories per day.
  1. Calculate the BMR for various animals whose masses are given in the table.
    Animal Bat Squirrel Raccoon Lynx Human Moose Rhinoceros
    Weight (kg) \(0.1\) \(0.6\) \(8\) \(30\) \(70\) \(360\) \(3500\)
    BMR (kcal/day) \(\) \(\) \(\) \(\) \(\) \(\) \(\)
  2. Sketch a graph of Kleiber’s rule for \(0 \lt m \le 400\text{.}\)
  3. Do larger species eat more or less, relative to their body mass, than smaller ones?
Solution.
  1. We evaluate the function for the values of \(m\) given. For example, to calculate the BMR of a bat, we compute
    \begin{equation*} B(0.1) = 70(0.1)^{0.75} = 12.4 \end{equation*}
    A bat expends, and hence must consume, at least \(12\) kilocalories per day. We evaluate the function to complete the rest of the table.
    Animal Bat Squirrel Raccoon Lynx Human Moose Rhinoceros
    Weight (kg) \(0.1\) \(0.6\) \(8\) \(30\) \(70\) \(360\) \(3500\)
    BMR (kcal/day) \(12\) \(48\) \(333\) \(897\) \(1694\) \(5785\) \(31,853\)
  2. We plot the data from the table to obtain the graph below.
    BMR vs mass
  3. If energy consumption were proportional to body weight, the graph would be a straight line. But because the exponent in Kleiber’s rule, \(\dfrac{3}{4}\text{,}\) is less than \(1\text{,}\) the graph is concave down, or bends downward. Therefore, larger species eat less than smaller ones, relative to their body weight.

Checkpoint 3.88. Practice 2.

  1. Complete the table of values for the function \(f (x) = x^{-3/4}\text{.}\)
    \(x\) \(0.1\) \(0.2\) \(0.5\) \(1\)
    \(f (x)\)
    \(x\) \(2\) \(5\) \(8\) \(10\)
    \(f (x)\)
  2. Sketch the graph of the function.
Answer 1.
\(5.62341\)
Answer 2.
\(3.3437\)
Answer 3.
\(1.68179\)
Answer 4.
\(1\)
Answer 5.
\(0.594604\)
Answer 6.
\(0.29907\)
Answer 7.
\(0.210224\)
Answer 8.
\(0.177828\)
Solution.
  1. \(x\) \(0.1\) \(0.2\) \(0.5\) \(1\)
    \(f (x)\) \(5.623\) \(3.344\) \(1.682\) \(1\)
\(x\) \(2\) \(5\) \(8\) \(10\)
\(f (x)\) \(0.595\) \(0.299\) \(0.210\) \(0.178\)
  1. A graph is below.
power function

Checkpoint 3.89. QuickCheck 2.

Describe the concavity of the graph of \(f(x)=x^{\frac{a}{b}}\text{,}\) where \(\dfrac{a}{b} \gt 0\text{.}\)
  • It is concave up.
  • It is concave down.
  • It is concave up if \(\dfrac{a}{b} \gt 1\text{,}\) and concave down if \(\dfrac{a}{b} \lt 1\text{.}\)
  • It is concave up if \(\dfrac{a}{b} \lt 1\text{,}\) and concave down if \(\dfrac{a}{b} \gt 1\text{.}\)
Answer.
\(\text{Choice 3}\)
Solution.
The graph is concave up if \(\dfrac{a}{b} \gt 1\text{,}\) and concave down if \(\dfrac{a}{b} \lt 1\text{.}\)

Subsection More about Scaling

In Example 3.87 we saw that large animals eat less than smaller ones, relative to their body weight. This is because the scaling exponent in Kleiber’s rule is less than \(1\text{.}\) For example, let \(s\) represent the mass of a squirrel. The mass of a moose is then \(600s\text{,}\) and its metabolic rate is
\begin{equation*} \begin{aligned}[t] B(600s) \amp = 70(600s)^{0.75} \\ \amp = 600^{0.75} \cdot 70s^{0.75} = 121B(s) \end{aligned} \end{equation*}
or \(121\) times the metabolic rate of the squirrel. Metabolic rate scales as \(k^{0.75}\text{,}\) compared to the mass of the animal.
In a famous experiment in the 1960s, an elephant was given LSD. The dose was determined from a previous experiment in which a \(2.6\)-kg cat was given \(0.26\) gram of LSD. Because the elephant weighed \(2970\) kg, the experimenters used a direct proportion to calculate the dose for the elephant:
\begin{equation*} \frac{0.26 \text{ g}}{2.6 \text{ kg}}= \frac{x \text{ g}}{2970 \text{ kg}} \end{equation*}
and arrived at the figure \(297\) g of LSD. Unfortunately, the elephant did not survive the experiment.

Example 3.90.

Use Kleiber’s rule and the dosage for a cat to estimate the corresponding dose for an elephant.
Solution.
If the experimenters had taken into account the scaling exponent of \(0.75\) in metabolic rate, they would have used a smaller dose. Because the elephant weighs \(\dfrac{2970}{2.6}\) , or about \(1142\) times as much as the cat, the dose would be \(1142^{0.75} = 196\) times the dosage for a cat, or about \(51\) grams.

Checkpoint 3.91. Practice 3.

A human being weighs about \(70\) kg, and \(0.2\) mg of LSD is enough to induce severe psychotic symptoms. Use these data and Kleiber’s rule to predict what dosage would produce a similar effect in an elephant.
About mg
Answer.
\(\left(\frac{2970}{70}\right)^{0.75}\cdot 0.2\)
Solution.
About \(3.3\) mg

Subsection Radical Notation

Because \(a^{1/n} = \sqrt[n]{a}\text{,}\) we can write any power with a fractional exponent in radical form as follows.

Rational Exponents and Radicals.

\begin{equation*} a^{m/n} = \sqrt[n]{a^m} =\left( \sqrt[n]{a}\right)^m \end{equation*}

Example 3.92.

  1. \(\displaystyle 125^{4/3} = \sqrt[3]{125^4} \text{ or } \left(\sqrt[3]{125}\right)^4\)
  2. \(\displaystyle x^{0.4} = x^{2/5} = \sqrt[5]{x^2}\)
  3. \(\displaystyle 6w^{-3/4} = \dfrac{6}{\sqrt[4]{w^3}}\)

Checkpoint 3.93. Practice 4.

Write each expression in radical notation.
  1. \(5t^{1.25}\)
    • \(\displaystyle \sqrt[4]{5t^{5}} \)
    • \(\displaystyle \sqrt[4]{(5t)^{5}} \)
    • \(\displaystyle 5\sqrt[4]{t^{5}} \)
    • \(\displaystyle 5\sqrt{(t/4)^{5}} \)
  2. \(3m^{-5/3}\)
    • \(\displaystyle \dfrac{3}{\sqrt[3]{m^{5}}} \)
    • \(\displaystyle \dfrac{1}{\sqrt[3]{3m^{5}}} \)
    • \(\displaystyle \dfrac{1}{3\sqrt[3]{m^{5}}} \)
    • \(\displaystyle -3 \sqrt[3]{m^{5}} \)
Answer 1.
\(\text{Choice 3}\)
Answer 2.
\(\text{Choice 1}\)
Solution.
  1. \(\displaystyle 5t^{1.25}=5\sqrt[4]{t^5}\)
  2. \(\displaystyle 3m^{-5/3}=\dfrac{3}{\sqrt[3]{m^5} }\)

Checkpoint 3.94. QuickCheck 3.

The notation \(a^{0.6}\) means
  • \(\displaystyle \sqrt[6]{a}\)
  • \(\displaystyle \dfrac{1}{a^6}\)
  • \(\displaystyle a^{\frac{1}{6}}\)
  • \(\displaystyle a^{\frac{3}{5}}\)
Answer.
\(\text{Choice 4}\)
Solution.
\(a^{0.6}=a^{\frac{3}{5}}\)
Usually, we will want to convert from radical notation to fractional exponents, since exponential notation is easier to use.

Example 3.95.

  1. \(\displaystyle \sqrt{x^5} = x^{5/2}\)
  2. \(\displaystyle 5 \sqrt[4]{p^3} = 5p^{3/4}\)
  3. \(\displaystyle \dfrac{3}{\sqrt[5]{t^2}}= 3t^{-2/5}\)
  4. \(\displaystyle \sqrt[3]{2y^2} = \left(2y^2\right)^{1/3} = 2^{1/3} y^{2/3}\)

Checkpoint 3.96. Practice 5.

Convert to exponential notation.
  1. \(\sqrt[3]{6w^2}=\)
  2. \(\sqrt[4]{\dfrac{v^3}{s^5}}=\)
Answer 1.
\(1.81712w^{\left({\frac{2}{3}}\right)}\)
Answer 2.
\(v^{\left({\frac{3}{4}}\right)}s^{\left(-{\frac{5}{4}}\right)}\)
Solution.
  1. \(\displaystyle 6^{1/3} w^{2/3}\)
  2. \(\displaystyle v^{3/4} s^{-5/4}\)

Subsection Operations with Rational Exponents

Powers with rational exponents—positive, negative, or zero—obey the laws of exponents, which we discussed in Section 3.1. You may want to review those laws before studying the following examples.

Example 3.97.

  1. \begin{equation*} \begin{aligned}[t] \frac{7^{0.75}}{7^{0.5}}\amp= 7^{0.75-0.5} = 7^{0.25} \amp\amp \blert{\text{Apply the second law of exponents.}} \end{aligned} \end{equation*}
  2. \begin{equation*} \begin{aligned}[t] v \cdot v^{-2/3}\amp= v^{1+(-2/3)} \amp\amp \blert{\text{Apply the first law of exponents.}}\\ \amp = v^{1/3} \end{aligned} \end{equation*}
  3. \begin{equation*} \begin{aligned}[t] \left(x^8\right)^{0.5}\amp= x^{8(0.5))} = x^4 \amp\amp \blert{\text{Apply the third law of exponents.}} \end{aligned} \end{equation*}
  4. \begin{equation*} \begin{aligned}[t] \frac{\left(5^{1/2}y^2\right)^2}{\left(5^{2/3} y\right)^3} \amp= \frac{5y^4}{5^2 y^3} \amp\amp \blert{\text{Apply the fourth law of exponents.}}\\ \amp = \frac{y^{4-3}}{5^{2-1}}=\frac{y}{5} \amp\amp \blert{\text{Apply the second law of exponents.}}\\ \end{aligned} \end{equation*}

Checkpoint 3.98. Practice 6.

Simplify by applying the laws of exponents.
  1. \(x^{1/3}\left(x + x^{2/3}\right)=\)
  2. \(\dfrac{n^{9/4}}{4n^{3/4}}=\)
Answer 1.
\(x^{\frac{4}{3}}+x\)
Answer 2.
\(\frac{n^{\frac{3}{2}}}{4}\)
Solution.
  1. \(\displaystyle x^{4/3}+x\)
  2. \(\displaystyle \dfrac{n^{3/2}}{4}\)

Checkpoint 3.99. QuickCheck 4.

Which of the following is the correct way to evaluate \(15^{\frac{3}{5}}\) on a calculator?
  • \(\displaystyle 15 \wedge 3 \div 5\)
  • \(\displaystyle 15 \div 3 \wedge 5\)
  • \(\displaystyle (15 \div 5) \wedge 3\)
  • \(\displaystyle 15 \wedge (3 \div 5)\)
Answer.
\(\text{Choice 4}\)
Solution.
\(15 \wedge (3 \div 5)\)

Subsection Solving Equations

According to the third law of exponents, when we raise a power to another power, we multiply the exponents together. In particular, if the two exponents are reciprocals, then their product is \(1\text{.}\) For example,
\begin{equation*} \left(x^{2/3}\right)^{3/2} = x^{(2/3) (3/2)} = x^1 = x \end{equation*}
This observation can help us to solve equations involving fractional exponents. For instance, to solve the equation
\begin{equation*} x^{2/3} = 4 \end{equation*}
we raise both sides of the equation to the reciprocal power, \(3/2\text{.}\) This gives us
\begin{equation*} \begin{aligned}[t] \left(x^{2/3}\right)^{3/2} \amp = 4^{3/2} \\ x \amp = 8 \end{aligned} \end{equation*}
The solution is \(8\text{.}\)

Example 3.100.

Solve \(~~(2x + 1)^{3/4} = 27\)
Solution.
We raise both sides of the equation to the reciprocal power, \(\dfrac{4}{3}\text{.}\)
\begin{equation*} \begin{aligned}[t] \left[(2x + 1)^{3/4}\right]^{4/3} \amp= 27^{4/3} \amp\amp \blert{\text{Apply the third law of exponents.}} \\ 2x + 1 \amp = 81 \amp\amp \blert{\text{Solve as usual.}} \\ x \amp = 40 \end{aligned} \end{equation*}

Checkpoint 3.101. QuickCheck 5.

To solve the equation \(x^{\frac{2}{3}}=k\) we should
  • A) Raise both sides to the negative of the exponent
  • B) Divide both sides by the exponent
  • C) Raise both sides to the reciprocal of the exponent
  • D) Raise the right side to the given exponent
Answer.
\(\text{C) ... the exponent}\)
Solution.
Raise both sides to the reciprocal of the exponent

Checkpoint 3.102. Practice 7.

Solve the equation \(3.2z^{0.6} - 9.7 = 8.7\text{.}\) Round your answer to two decimal places.
Answer:
Hint.
\(\blert{\text{Isolate the power.}}\)
\(\blert{\text{Raise both sides to the reciprocal power}}.\)
Answer.
\(\left(\frac{18.4}{3.2}\right)^{\frac{1}{0.6}}\)
Solution.
\(18.45\)

Checkpoint 3.103. Pause and Reflect.

Explain why \(x\sqrt{x}=x^{1.5}\text{.}\)

Investigation 3.2. Vampire Bats.

Small animals such as bats cannot survive for long without eating. The graph below shows how the weight, \(W\text{,}\) of a typical vampire bat decreases over time until its next meal, until the bat reaches the point of starvation. The curve is the graph of the function
\begin{equation*} W(h) = 130.25h^{-0.126} \end{equation*}
where \(h\) is the number of hours since the bat’s most recent meal. (Source: Wilkinson, 1984)
bat weight
  1. Use the graph to estimate answers to the following questions: How long can the bat survive after eating until its next meal? What is the bat’s weight at the point of starvation?
  2. Use the formula for \(W(h)\) to verify your answers.
  3. Write and solve an equation to answer the question: When the bat’s weight has dropped to \(90\) grams, how long can it survive before eating again?
  4. Complete the table showing the number of hours since the bat last ate when its weight has dropped to the given values.
    Weight (grams) \(97.5\) \(92.5\) \(85\) \(80\)
    Hours since eating \(\) \(\) \(\) \(\)
    Point on graph \(A\) \(B\) \(C\) \(D\)
  5. Compute the slope of the line segments from point \(A\) to point \(B\text{,}\) and from point \(C\) to point \(D\text{.}\) Include units in your answers.
    bat weight
  6. What happens to the slope of the curve as \(h\) increases? What does this tell you about the concavity of the curve?
  7. Suppose a bat that weighs \(80\) grams consumes \(5\) grams of blood. How many hours of life does it gain? Suppose a bat that weighs \(97.5\) grams gives up a meal of \(5\) grams of blood. How many hours of life does it forfeit?
  8. Vampire bats sometimes donate blood (through regurgitation) to other bats that are close to starvation. Suppose a bat at point \(A\) on the curve donates \(5\) grams of blood to a bat at point \(D\text{.}\) Explain why this strategy is effective for the survival of the bat community.

Subsection Section Summary

Subsubsection Vocabulary

Look up the definitions of new terms in the Glossary.
  • Rational exponent

Subsubsection CONCEPTS

  1. Rational exponents: \(a^{m/n} = \left(a^{1/n}\right)^m = \left(a^m\right)^{1/n},~~~ a\gt 0,~~ n \ne 0.\)
  2. To compute \(a^{m/n}\text{,}\) we can compute the \(n\)th root first, or the \(m\)th power, whichever is easier.
  3. The graphs of power functions \(y = x^{m/n}\text{,}\) where \(m/n\) is positive, are all increasing for \(x\ge 0\text{.}\) If \(m/n\gt 1\text{,}\) the graph is concave up. If \(0 \lt m/n \lt 1\text{,}\) the graph is concave down.
  4. Radical notation: \(a^{m/n} =\sqrt[n]{a^m} = \left( \sqrt[n]{a}\right)^m\text{.}\)
  5. Powers with rational exponents—positive, negative, or zero—obey the laws of exponents.
  6. To solve the equation \(x^{m/n} = k\text{,}\) we raise both sides to the power \(n/m\text{.}\)

Subsubsection STUDY QUESTIONS

  1. What does the notation \(a^{0.98}\) mean?
  2. Explain how to evaluate the function \(f (x) = x^{-3/4}\) for \(x = 625\text{,}\) without using a calculator.
  3. Explain why \(x \sqrt{x}=x^{1.5} \text{.}\)
  4. What is the first step in solving the equation \((x - 2)^{-5/2} = 1.8\text{?}\)
  5. If the graph of \(f (x) = x^{a/b}\) is concave down, and \(a/b \gt 0\text{,}\) what else can you say about \(a/b\text{?}\)

Subsubsection SKILLS

Practice each skill in the Homework problems listed.
  1. Simplify and evaluate powers with rational exponents: #1–4, 13–18
  2. Graph power functions with rational exponents: #19–22
  3. Solve radical equations: #23–38, 59 and 60
  4. Analyze power functions with rational exponents: #23–36
  5. Simplify expressions using the laws of exponents: #37–44, 57–70
  6. Solve equations involving rational exponents: #45–56

Exercises Homework 3.4

For the problems in Homework 3.4, assume that all variables represent positive numbers.

Exercise Group.

For Problems 1-4, evaluate each power.
1.
  1. \(\displaystyle 81^{3/4} \)
  2. \(\displaystyle 125^{2/3} \)
  3. \(\displaystyle 625^{0.75} \)
2.
  1. \(\displaystyle -8^{2/3} \)
  2. \(\displaystyle -64^{2/3} \)
  3. \(\displaystyle 243^{0.4} \)
3.
  1. \(\displaystyle 16^{-3/2} \)
  2. \(\displaystyle 8^{-4/3} \)
  3. \(\displaystyle 32^{-1.6} \)
4.
  1. \(\displaystyle -125^{-4/3} \)
  2. \(\displaystyle -32^{-3/5} \)
  3. \(\displaystyle 100^{-2.5} \)

Exercise Group.

For Problems 5–8, write each power in radical form.
5.
  1. \(\displaystyle x^{4/5} \)
  2. \(\displaystyle b^{-5/6} \)
  3. \(\displaystyle (pq)^{-2/3} \)
6.
  1. \(\displaystyle y^{3/4} \)
  2. \(\displaystyle a^{-2/7} \)
  3. \(\displaystyle (st)^{-3/5} \)
7.
  1. \(\displaystyle 3x^{0.4} \)
  2. \(\displaystyle 4z^{-4/3} \)
  3. \(\displaystyle -2x^{0.25}y^{0.75} \)
8.
  1. \(\displaystyle 5y^{2/3} \)
  2. \(\displaystyle 6w^{-1.5} \)
  3. \(\displaystyle -3x^{0.4}y^{0.6} \)

Exercise Group.

For Problems 9–12, write each expression with fractional exponents.
9.
  1. \(\displaystyle \sqrt[3]{x^2} \)
  2. \(\displaystyle 2\sqrt[5]{ab^3} \)
  3. \(\displaystyle \dfrac{-4m}{\sqrt[6]{p^7}} \)
10.
  1. \(\displaystyle \sqrt{y^3} \)
  2. \(\displaystyle 6\sqrt[5]{(ab)^3} \)
  3. \(\displaystyle \dfrac{-2n}{\sqrt[8]{q^{11}}} \)
11.
  1. \(\displaystyle \sqrt[3]{(ab)^{2}} \)
  2. \(\displaystyle \dfrac{8}{\sqrt[4]{x^3}} \)
  3. \(\displaystyle \dfrac{R}{3\sqrt{TK^5}} \)
12.
  1. \(\displaystyle \sqrt[3]{ab^2} \)
  2. \(\displaystyle \dfrac{5}{\sqrt[3]{y^2}} \)
  3. \(\displaystyle \dfrac{S}{4\sqrt{VH^3}} \)

Exercise Group.

For Problems 13–16, evaluate each root without using a calculator.
13.
  1. \(\displaystyle \sqrt[5]{32^3} \)
  2. \(\displaystyle -\sqrt[3]{27^4} \)
  3. \(\displaystyle \sqrt[4]{16y^{12}} \)
14.
  1. \(\displaystyle \sqrt[4]{16^5} \)
  2. \(\displaystyle -\sqrt[3]{125^2} \)
  3. \(\displaystyle \sqrt[5]{243x^{10}} \)
15.
  1. \(\displaystyle -\sqrt{a^8b^{16}} \)
  2. \(\displaystyle \sqrt[3]{8x^9y^{27}} \)
  3. \(\displaystyle -\sqrt[4]{81a^8b^{12}} \)
16.
  1. \(\displaystyle -\sqrt{a^{10} b^{36}} \)
  2. \(\displaystyle \sqrt[3]{64x^6 y^{18}} \)
  3. \(\displaystyle -\sqrt[5]{32x^{25}y^{5}} \)

Exercise Group.

For Problems 17–18, use a calculator to approximate each power or root to the nearest thousandth.
17.
  1. \(\displaystyle 12^{5/6} \)
  2. \(\displaystyle \sqrt[3]{6^4} \)
  3. \(\displaystyle 37^{-2/3} \)
  4. \(\displaystyle 4.7^{2.3} \)
18.
  1. \(\displaystyle 20^{5/4} \)
  2. \(\displaystyle \sqrt[5]{8^3} \)
  3. \(\displaystyle 128^{-3/4} \)
  4. \(\displaystyle 16.1^{0.29} \)

19.

During a flu epidemic in a small town, health officials estimate that the number of people infected \(t\) days after the first case was discovered is given by
\begin{equation*} I (t) = 50 t^{3/5} \end{equation*}
  1. Make a table of values for \(I (t)\) on the domain \(0\le t\le 20\text{.}\) What is the range of the function on that domain?
    \(t\) \(5\) \(10\) \(15\) \(20\)
    \(I(t)\) \(\hphantom{00000} \) \(\hphantom{00000} \) \(\hphantom{00000} \) \(\hphantom{00000} \)
  2. How long will it be before \(300\) people are ill?
  3. Graph the function \(I(t)\) and verify your answer to part (b) on your graph.

20.

The research division of an advertising firm estimates that the number of people who have seen their ads \(t\) days after the campaign begins is given by the function
\begin{equation*} N (t) = 2000 t^{5/4} \end{equation*}
  1. Make a table of values for \(N (t)\) on the domain \(0\le t\le 20\text{.}\) What is the range of the function on that domain?
    \(t\) \(6\) \(10\) \(14\) \(20\)
    \(N(t)\) \(\hphantom{00000} \) \(\hphantom{00000} \) \(\hphantom{00000} \) \(\hphantom{00000} \)
  2. How long will it be before \(75,000\) people have seen the ads?
  3. Graph the function \(N(t)\) and verify your answer to part (b) on your graph.

Exercise Group.

In Problems 21–22, graph each set of power functions in the suggested window and compare the graphs.
21.
\(y_1 = x,~~ y_2 = x^{5/4},~~ y_3 = x^{3/2},~~ y_4 = x^2,~~ y_5 = x^{5/2}\)
\({\text{Xmin}} = 0,~~~ {\text{Xmax}} = 6,~~~ {\text{Ymin}} = 0,~~~ {\text{Ymax}} = 10\)
22.
\(y_1 = x^{2/5},~~ y_2 = x^{1/2},~~ y_3 = x^{2/3},~~ y_4 = x^{3/4},~~ y_5 = x\)
\({\text{Xmin}} = 0,~~~ {\text{Xmax}} = 6,~~~ {\text{Ymin}} = 0,~~~ {\text{Ymax}} = 4\)

23.

The surface to volume ratio is important in studying how organisms grow and why animals of different sizes have different characteristics.
  1. Write formulas for the volume, \(V\text{,}\) and the surface area, \(A\text{,}\) of a cube in terms of its length, \(L\text{.}\)
  2. Express the length of the cube as a function of its volume. Express the length of the cube as a function of its surface area.
  3. Express the surface area of the cube as a function of its volume.
  4. Express the surface to volume ratio of a cube in terms of its length. What happens to the surface to volume ratio as \(L\) increases?

24.

Repeat Problem 23 for the volume and surface area of a sphere in terms of its radius, \(R\text{.}\)
  1. Write formulas for the volume, \(V\text{,}\) and the surface area, \(A\text{,}\) of a sphere in terms of its radius, \(R\text{.}\)
  2. Express the radius of the sphere as a function of its volume. Express the radius of the sphere as a function of its surface area.
  3. Express the surface area of the sphere as a function of its volume.
  4. Express the surface to volume ratio of a sphere in terms of its radius. What happens to the surface to volume ratio as \(R\) increases?

25.

A brewery wants to replace its old vats with larger ones. To estimate the cost of the new equipment, the accountant uses the \(0.6\) rule for industrial costs, which states that the cost of a new container is approximately \(N = Cr^{0.6}\text{,}\) where \(C\) is the cost of the old container and \(r\) is the ratio of the capacity of the new container to the old one.
  1. If an old vat cost $\(5000\text{,}\) graph \(N\) as a function of \(r\text{.}\)
  2. How much should the accountant budget for a new vat that holds \(1.8\) times as much as the old one?

26.

If a quantity of air expands without changing temperature, its pressure, in pounds per square inch, is given by \(P = kV^{-1.4}\text{,}\) where \(V\) is the volume of the air in cubic inches and \(k = 2.79\times 10^4\text{.}\)
  1. Graph \(P\) as a function of \(V\text{.}\)
  2. Find the air pressure of an air sample when its volume is \(50\) cubic inches.

27.

In the 1970s, Jared Diamond studied the number of bird species on small islands near New Guinea. He found that larger islands support a larger number of different species, according to the formula
\begin{equation*} S = 15.1A^{0.22} \end{equation*}
where \(S\) is the number of species on an island of area \(A\) square kilometers. (Source: Chapman and Reiss, 1992)
  1. Fill in the table.
    \(A\) \(10\) \(100\) \(1000\) \(5000\) \(10,000\)
    \(S\) \(\hphantom{000000} \) \(\hphantom{000000} \) \(\hphantom{000000} \) \(\hphantom{000000} \) \(\hphantom{000000} \)
  2. Graph the function on the domain \(0\lt A\le 10,000\text{.}\)
  3. How many species of birds would you expect to find on Manus Island, with an area of \(2100\) square kilometers? On Lavongai, whose area is \(1140\) square kilometers?
  4. How large must an island be in order to support \(200\) different species of bird?

28.

The drainage basin of a river channel is the area of land that contributes water to the river. The table gives the lengths in miles of some of the world’s largest rivers and the areas of their drainage basins in square miles. (Source: Leopold, Wolman, and Miller 1992)
  1. Plot the data, using units of \(100,000\) on the horizontal axis and units of \(500\) on the vertical axis.
  2. The length, \(L\text{,}\) of the channel is related to the area, \(A\text{,}\) of its drainage basin according to the formula
    \begin{equation*} L = 1.05A^{0.58} \end{equation*}
    Graph this function on top of the data points.
  3. The drainage basin for the Congo covers about \(1,600,000\) square miles. Estimate the length of the Congo River.
  4. The Rio Grande is \(1700\) miles long. What is the area of its drainage basin?
River Area of
drainage basin		 Length
Amazon \(2,700,000\) \(4300\)
Nile \(1,400,000\) \(4200\)
Mississippi \(1,300,000\) \(4100\)
Yangtze \(580,000\) \(2900\)
Volga \(480,000\) \(2300\)
St. Lawrence \(460,000\) \(1900\)
Ganges \(440,000\) \(1400\)
Orinoco \(380,000\) \(1400\)
Indus \(360,000\) \(2000\)
Danube \(350,000\) \(1800\)
Colorado \(250,000\) \(1700\)
Platte \(72,000\) \(800\)
Rhine \(63,000\) \(900\)
Seine \(48,000\) \(500\)
Delaware \(12,000\) \(200\)

29.

The table at right shows the exponent, \(p\text{,}\) in the allometric equation
\begin{gather*} \text{variable} = k (\text{body mass})^p \end{gather*}
for some variables related to mammals. (Source: Chapman and Reiss, 1992)
Variable Exponent, \(p\)
Home range size \(1.26\)
Lung volume \(1.02\)
Brain mass \(0.70\)
Respiration rate \(-0.26\)
  1. Match each equation to one of the graphs shown in the figure.
    four power functions
  2. Explain how the value of \(p\) in the allometric equation determines the shape of the graph. Consider the cases \(p\gt 1\text{,}\) \(0\lt p\lt 1\text{,}\) and \(p\lt 0\text{.}\)

30.

The average body mass of a dolphin is about \(140\) kilograms, twice the body mass of an average human male.
  1. Using the allometric equations in Problem 29, calculate the ratio of the brain mass of a dolphin to that of a human.
  2. A good-sized brown bear weighs about \(280\) kilograms, twice the weight of a dolphin. Calculate the ratio of the brain mass of a brown bear to that of a dolphin.
  3. Use a ratio to compare the heartbeat frequencies of a dolphin and a human, and those of a brown bear and a dolphin. (See Example 3.70 of Section 3.3.)

31.

The gourd species Tricosanthes grows according to the formula \(L = ad^{2.2}\text{,}\) where \(L\) is its length and \(d\) is its width. The species Lagenaria has the growth law \(L = ad^{0.81}\text{.}\) (Source: Burton, 1998)
  1. By comparing the exponents, predict which gourd grows into a long, thin shape, and which is relatively fatter. Which species is called the snake gourd, and which is the bottle gourd?
  2. The snake gourd reaches a length of \(2\) meters (\(200\) cm), with a diameter of only \(4\) cm. Find the value of \(a\) in its growth law.
  3. The bottle gourd is \(10\) cm long and \(7\) cm in diameter at maturity. Find the value of \(a\) in its growth law.
  4. The giant bottle gourd grows to a length of \(23\) cm with a diameter of \(20\) cm. Does it grow according to the same law as standard bottle gourds?

32.

As a fiddler crab grows, one claw (called the chela) grows much faster than the rest of the body. The table shows the mass of the chela, \(C\text{,}\) versus the mass of the rest of the body, \(b\text{,}\) for a number of fiddler crabs. (Source: Burton, 1998)
\(b\) \(65\) \(110\) \(170\) \(205\) \(300\) \(360\) \(615\)
\(C\) \(6\) \(15\) \(30\) \(40\) \(68\) \(110\) \(240\)
  1. Plot the data.
  2. On the same axes, graph the function \(C = 0.007b^{1.63}\text{.}\) How well does the function fit the data?
  3. Using the function in part (b), predict the chela mass of a fiddler crab if the rest of its body weighs \(400\) mg.
  4. The chela from a fiddler crab weighs \(250\) mg. How much does the rest of its body weigh?
  5. As the body mass of a fiddler crab doubles from \(100\) mg to \(200\) mg, by what factor does the mass of its chela increase? As the body mass doubles from \(200\) mg to \(400\) mg?

33.

The climate of a region has a great influence on the types of animals that can survive there. Extreme temperatures create difficult living conditions, so the diversity of wildlife decreases as the annual temperature range increases. Along the west coast of North America, the number of species of mammals, \(M\text{,}\) is approximately related to the temperature range, \(R\text{,}\) (in degrees Celsius) by the function \(M = f (R) = 433.8R^{-0.742}\text{.}\) (Source: Chapman and Reiss, 1992)
  1. Graph the function for temperature ranges up to \(30\degree\)C.
  2. How many species would you expect to find in a region where the temperature range is \(10\degree \)C? Label the corresponding point on your graph.
  3. If \(50\) different species are found in a certain region, what temperature range would you expect the region to experience? Label the corresponding point on your graph.
  4. Evaluate the function to find \(f (9)\text{,}\) \(f (10)\text{,}\) \(f (19)\text{,}\) and \(f (20)\text{.}\) What do these values represent? Calculate the change in the number of species as the temperature range increases from \(9\degree\)C to \(10\degree\)C and from \(19\degree\)C to \(20\degree\)C. Which \(1\degree\) increase results in a greater decrease in diversity? Explain your answer in terms of slopes on your graph.

34.

A bicycle ergometer is used to measure the amount of power generated by a cyclist. The scatterplot shows how long an athlete was able to sustain various levels of power output. The curve is the graph of \(y = 500x^{-0.29}\text{,}\) which approximately models the data. (Source: Alexander, 1992)
scatterplot and power function
  1. In this graph, which variable is independent and which is dependent?
  2. The athlete maintained \(650\) watts of power for \(40\) seconds. What power output does the equation predict for \(40\) seconds?
  3. The athlete maintained \(300\) watts of power for \(10\) minutes. How long does the equation predict that power output can be maintained?
  4. In 1979, a remarkable pedal-powered aircraft called the Gossamer Albatross was successfully flown across the English Channel. The flight took \(3\) hours. According to the equation, what level of power can be maintained for \(3\) hours?
  5. The Gossamer Albatross needed \(250\) watts of power to keep it airborne. For how long can \(250\) watts be maintained according to the given equation?

35.

Investigation 3.1 at the start of this chapter gives data for the pressure inside April and Tolu’s balloon as a function of its diameter. As the diameter of the balloon increases from \(5\) cm to \(20\) cm, the pressure inside decreases. Can we find a function that describes this portion of the graph?
  1. Pressure is the force per unit area exerted by the balloon on the air inside, or \(P = \dfrac{F}{A}\text{.}\) Because the balloon is spherical, its surface area, \(A\text{,}\) is given by \(A = \pi d^2\text{.}\) Because the force increases as the balloon expands, we will try a power function \(F = kd^p\text{,}\) where \(k\) and \(p\) are constants, and see if this fits the data. Combine the three equations, \(P = \dfrac{F}{A}\text{,}\) \(A = \pi d^2\text{,}\) and \(F = kd^p\text{,}\) to express \(P\) as a power function of \(d\text{.}\)
  2. Use your calculator’s power regression feature to find a power function that fits the data. Graph the function \(P = 211d^{-0.7}\) on top of the data. Do the data support the hypothesis that \(P\) is a power function of \(d\text{?}\)
  3. What is the value of the exponent \(p\) in \(F = kd^p\text{?}\)

36.

The table shows the total number of frequent flyer miles redeemed by customers through the given year. (Source: www.hotelnewsresource.com)
  1. Plot the data, with \(t = 0\) in 1980. What type of function might model the data?
  2. Graph the function \(f (t) = 3.13t^{2.33}\) on top of the data.
  3. Evaluate \(f (5)\) and \(f (25)\text{.}\) What do those values mean in this context?
  4. Use the regression equation to predict when the total number of miles redeemed will reach 10 trillion.
Year Cumulative
miles redeemed
(billions)
\(1982\) \(14.8\)
\(1984\) \(85.3\)
\(1986\) \(215.4\)
\(1988\) \(387.5\)
\(1990\) \(641.3\)
\(1992\) \(975.2\)
\(1994\) \(1455.9\)
\(1996\) \(1996\)
\(1998\) \(2670.8\)
\(2000\) \(3379.1\)
\(2002\) \(4123.6\)
Hint.
(For part d): How many billions make a trillion?

Exercise Group.

For Problems 37–42, simplify by applying the laws of exponents. Write your answers with positive exponents only.
37.
  1. \(\displaystyle 4a^{6/5}a^{4/5} \)
  2. \(\displaystyle 9b^{4/3}b^{1/3}\)
38.
  1. \(\displaystyle (-2m^{2/3})^4 \)
  2. \(\displaystyle (-5n^{3/4})^3\)
39.
  1. \(\displaystyle \dfrac{8w^{9/4}}{2w^{3/4}} \)
  2. \(\displaystyle \dfrac{12z^{11/3}}{4z^{5/3}}\)
40.
  1. \(\displaystyle (-3u^{5/3}) (5u^{-2/3}) \)
  2. \(\displaystyle (-2v^{7/8})(-3v^{-3/8})\)
41.
  1. \(\displaystyle \dfrac{k^{3/4}}{2k} \)
  2. \(\displaystyle \dfrac{4h^{2/3}}{3h}\)
42.
  1. \(\displaystyle c^{-2/3}\left(\dfrac{2}{3}c^2 \right) \)
  2. \(\displaystyle \dfrac{r^3}{4}(r^{-5/2}) \)

43.

The incubation time for a bird’s egg is a function of the mass, \(m\text{,}\) of the egg, and has been experimentally determined as
\begin{equation*} I(m) = 12.0m^{0.217} \end{equation*}
where \(m\) is measured in grams and \(I\) is in days. (Source: Burton, 1998)
  1. Calculate the incubation time (to the nearest day) for the wren, whose eggs weigh about \(2.5\) grams, and the greylag goose, whose eggs weigh \(46\) grams.
  2. During incubation, birds’ eggs lose water vapor through their porous shells. The rate of water loss from the egg is also a function of its mass, and it appears to follow the rule
    \begin{equation*} W(m) = 0.015m^{0.742} \end{equation*}
    in grams per day. Combine the functions \(I(m)\) and \(W(m)\) to calculate the fraction of the initial egg mass that is lost during the entire incubation period.
  3. Explain why your result shows that most eggs lose about \(18\%\) of their mass during incubation.

44.

The incubation time for birds’ eggs is given by
\begin{equation*} I(m) = 12.0m^{0.217} \end{equation*}
where \(m\) is the weight of the egg in grams, and \(I\) is in days. (See Problem 43.) Before hatching, the eggs take in oxygen at the rate of
\begin{equation*} O(m) = 22.2m^{0.77} \end{equation*}
in milliliters per day. (Source: Burton, 1998)
  1. Combine the functions \(I(m)\) and \(O(m)\) to calculate the total amount of oxygen taken in by the egg during its incubation.
  2. Use your result from part (a) to explain why total oxygen consumption per unit mass is approximately inversely proportional to incubation time.
  3. Predict the oxygen consumption per gram of a herring gull’s eggs, given that their incubation time is \(26\) days. (The actual value is \(11\) milliliters per day.)

Exercise Group.

For Problems 45–50, solve. Round your answers to the nearest thousandth if necessary.
45.
\(x^{2/3} - 1 = 15\)
46.
\(x^{3/4} + 3 = 11 \)
47.
\(x^{-2/5} = 9\)
48.
\(x^{-3/2} = 8 \)
49.
\(2(5.2 - x^{5/3}) = 1.4\)
50.
\(3(8.6 - x^{5/2}) = 6.5 \)

51.

Kepler’s law gives a relation between the period, \(p\text{,}\) of a planet’s revolution, in years, and its average distance, \(a\text{,}\) from the sun:
\begin{equation*} p^2 = Ka^3 \end{equation*}
where \(K = 1.243\times 10^{-24}\text{,}\) \(a\) is measured in miles, and \(p\) is in years.
  1. Solve Kepler’s law for \(p\) as a function of \(a\text{.}\)
  2. Find the period of Mars if its average distance from the sun is \(1.417\times 10^8\) miles.

52.

Refer to Kepler’s law, \(p^2 = Ka^3\text{,}\) in Problem 51.
  1. Solve Kepler’s law for \(a\) as a function of \(p\text{.}\)
  2. Find the distance from Venus to the sun if its period is \(0.615\) years.

53.

If \(f (x) = (3x - 4)^{3/2}\text{,}\) find \(x\) so that \(f (x) = 27\text{.}\)

54.

If \(g(x) = (6x - 2)^{5/3}\text{,}\) find \(x\) so that \(g(x) = 32\text{.}\)

55.

If \(S(x) = 12x^{-5/4}\text{,}\) find \(x\) so that \(S(x) = 20\text{.}\)

56.

If \(T (x) = 9x^{-6/5}\text{,}\) find \(x\) so that \(T (x) = 15\text{.}\)

Exercise Group.

For Problems 57–64, use the distributive law to find the product.
57.
\(2x^{1/2}(x-x^{1/2}) \)
58.
\(x^{1/3} (2x^{2/3} - x^{1/3})\)
59.
\(\dfrac{1}{2}y^{-1/3}(y^{2/3}+3y^{-5/6}) \)
60.
\(3y^{-3/8}\left(\dfrac{1}{4}y^{-1/4} + y^{3/4} \right)\)
61.
\((2x^{1/4} + 1) (x^{1/4} - 1) \)
62.
\((2x^{1/3} - 1) (x^{1/3} + 1)\)
63.
\((a^{3/4}-2)^2 \)
64.
\((a^{2/3}+3)^2\)

Exercise Group.

For Problems 65–70, factor out the smallest power from each expression. Write your answers with positive exponents only.
65.
\(x^{3/2} + x = x(~~\text{?}~~)\)
66.
\(y - y^{2/3} = y^{2/3}(~~\text{?}~~)\)
67.
\(y^{3/4} - y^{-1/4} = y^{-1/4}(~~\text{?}~~)\)
68.
\(x^{-3/2} + x^{-1/2} = x^{-3/2}(~~\text{?}~~)\)
69.
\(a^{1/3} + 3 - a^{-1/3} = a^{-1/3}(~~\text{?}~~)\)
70.
\(3b - b^{3/4} + 4b^{-3/4} = b^{-3/4}(~~\text{?}~~)\)
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