Subsection \(n\)th Roots
Recall that \(s\) is a square root of \(b\) if \(s^2 = b\text{,}\) and \(s\) is a cube root of \(b\) if \(s^3 = b\text{.}\) In a similar way, we can define the fourth, fifth, or sixth root of a number. For instance, the fourth root of \(b\) is a number \(s\) whose fourth power is \(b\text{.}\) In general, we make the following definition.
\(n\)th Roots.
\(s\) is called an \(n\)th root of \(b\) if \(s^n = b\text{.}\)
We use the symbol \(\sqrt[n]{b}\) to denote the \(n\)th root of \(b\text{.}\) An expression of the form \(\sqrt[n]{b}\) is called a radical, \(b\) is called the radicand, and \(n\) is called the index of the radical.
Example 3.50.
\(\sqrt[4]{81} = 3\) because \(3^4 = 81\)
\(\sqrt[5]{32} = 2\) because \(2^5 = 32\)
\(\sqrt[6]{64} = 2\) because \(2^6 = 64\)
\(\sqrt[4]{1} = 1\) because \(1^4 = 1\)
\(\sqrt[5]{100,000} = 10\) because \(10^5 = 100,000\)
Checkpoint 3.51. Practice 1.
Evaluate each radical.
\(\displaystyle \sqrt[4]{16}\)
\(\displaystyle \sqrt[5]{243}\)
Solution.
\(\displaystyle 2\)
\(\displaystyle 3\)
Subsection Exponential Notation for Radicals
A convenient notation for radicals uses fractional exponents. Consider the expression \(9^{1/2}\text{.}\) What meaning can we attach to an exponent that is a fraction? The third law of exponents says that when we raise a power to a power, we multiply the exponents together:
\begin{equation*}
\left(x^a\right)^b = x^{ab}
\end{equation*}
Therefore, if we square the number \(9^{1/2}\text{,}\) we get
\begin{equation*}
\left(9^{1/2}\right)^2 = 9^{(1/2)(2)} = 9^1 = 9
\end{equation*}
Thus, \(9^{1/2}\) is a number whose square is \(9\text{.}\) But this means that \(9^{1/2}\) is a square root of \(9\text{,}\) or
\begin{equation*}
9^{1/2} =\sqrt{9} = 3
\end{equation*}
In general, any nonnegative number raised to the \(1/2\) power is equal to the positive square root of the number, or
\begin{equation*}
a^{1/2} =\sqrt{a}
\end{equation*}
Example 3.52.
\(\displaystyle 25^{1/2} = 5\)
\(\displaystyle -25^{1/2} = -5\)
\((-25)^{1/2}\) is not a real number.
\(\displaystyle 0^{1/2} = 0\)
Checkpoint 3.53. Practice 2.
Evaluate each power.
\(\displaystyle 4^{1/2}\)
\(\displaystyle 4^{-2}\)
\(\displaystyle 4^{-1/2}\)
\(\displaystyle \left(\dfrac{1}{4}\right)^{1/2}\)
Solution.
\(\displaystyle 2\)
\(\displaystyle \dfrac{1}{16} \)
\(\displaystyle \dfrac{1}{2} \)
\(\displaystyle \dfrac{1}{2} \)
The same reasoning works for roots with any index. For instance, \(8^{1/3}\) is the cube root of \(8\text{,}\) because
\begin{equation*}
\left(8^{1/3}\right)^3 = 8^{(1/3)(3)} = 8^1 = 8
\end{equation*}
In general, we make the following definition for fractional exponents.
Exponential Notation for Radicals.
For any integer \(n \ge 2\) and for \(a \ge 0\text{,}\)
\begin{equation*}
a^{1/n} = \sqrt[n]{a}
\end{equation*}
Example 3.54.
\(\displaystyle 81^{1/4} = \sqrt[4]{81} = 3\)
\(\displaystyle 125^{1/3} = \sqrt[3]{125} = 5\)
Caution 3.55.
Note that
\begin{equation*}
25^{1/2} \ne \frac{1}{2}(25) ~~ \text{ and } ~~ 125^{1/3} \ne \frac{1}{3}(125)
\end{equation*}
An exponent of \(\dfrac{1}{2}\) denotes the square root of its base, and an exponent of \(\dfrac{1}{3}\) denotes the cube root of its base.
Checkpoint 3.56. Practice 3.
Write each power with radical notation, and then evaluate.
\(\displaystyle 32^{1/5} = \fillinmath{XXXX} = \fillinmath{XXXX}\)
\(\displaystyle 625^{1/4}= \fillinmath{XXXX}= \fillinmath{XXXX}\)
Solution.
\(\displaystyle 32^{1/5}=\sqrt[5]{32}= 2\)
\(\displaystyle 625^{1/4}=\sqrt[4]{625}= 5\)
Of course, we can use decimal fractions for exponents as well. For example,
\begin{equation*}
\sqrt{a} = a^{1/2} = a^{0.5} ~~~~\text{ and } ~~~~ \sqrt[4]{a} = a^{1/4} = a^{0.25}
\end{equation*}
Checkpoint 3.57. QuickCheck 1.
Which of the following expressions is not equivalent to the other three?
\(\displaystyle B^{\frac{1}{8}}\)
\(\displaystyle \sqrt[8]{B}\)
\(\displaystyle B^{0.125}\)
\(\displaystyle \dfrac{1}{B^8}\)
Solution.
\(\dfrac{1}{B^8}\) is not equivalent to the other three expressions.
Example 3.58.
\(\displaystyle 100^{0.5} = \sqrt{100} = 10\)
\(\displaystyle 16^{0.25} = \sqrt[4]{16} = 2\)
Checkpoint 3.59. Practice 4.
Write each power with radical notation, and then evaluate.
\(\displaystyle 100,000^{0.2} = \fillinmath{XXXX} = \fillinmath{XXXX}\)
\(\displaystyle 81^{0.25}= \fillinmath{XXXX} = \fillinmath{XXXX}\)
Solution.
\(\displaystyle 100,000^{0.2}=\sqrt[5]{100,000}=10 \)
\(\displaystyle 81^{0.25}=\sqrt[4]{81}=3 \)
Checkpoint 3.60. Pause and Reflect.
Explain why \(x^{\frac{1}{4}}\) is a reasonable notation for \(\sqrt[4]{x}\text{.}\)
Subsection Irrational Numbers
What about \(n\)th roots such as \(\sqrt{23}\) and \(5^{1/3}\) that cannot be evaluated easily? These are examples of irrational numbers. We can use a calculator to obtain decimal approximations for irrational numbers. For example, you can verify that
\begin{equation*}
\sqrt{23} \approx 4.796 ~ \text{ and } ~ 5^{1/3}\approx 1.710
\end{equation*}
It is not possible to write down an exact decimal equivalent for an irrational number, but we can find an approximation to as many decimal places as we like.
Caution 3.61.
The following keying sequence for evaluating the irrational number \(7^{1/5}\) is incorrect:
7 ^ 1 ÷ 5 ENTER
You can check that this sequence calculates \(\dfrac{7^1}{5}\text{,}\) instead of \(7^{1/5}\text{.}\) Recall that according to the order of operations, powers are computed before multiplications or divisions. We must enclose the exponent \(1/5\) in parentheses and enter
7 ^ ( 1 ÷ 5 ) ENTER
Or, because \(\frac{1}{5}= 0.2\text{,}\) we can enter
7 ^ 0.2 ENTER
Subsection Using Fractional Exponents to Solve Equations
In Chapter 2, we learned that raising to powers and taking roots are inverse operations, that is, each operation undoes the effects of the other. This relationship is especially easy to see when the root is denoted by a fractional exponent. For example, to solve the equation
\begin{equation*}
x^4 = 250
\end{equation*}
we would take the fourth root of each side. But instead of using radical notation, we can raise both sides of the equation to the power \(\dfrac{1}{4}\text{:}\)
\begin{equation*}
\left(x^4\right)^{1/4} = 250^{1/4}
\end{equation*}
The third law of exponents tells us that \(\left(x^a\right)^b = x^{ab}\text{,}\) so
\begin{equation*}
\left(x^4\right)^{1/4} = x^{(1/4)(4)} = x^1
\end{equation*}
We evaluate the right side to find \(250^{1/4}\) or \(\sqrt[4]{250} \approx 3.98\text{.}\)
In general, to solve an equation involving a power function \(x^n\text{,}\) we first isolate the power, then raise both sides to the exponent \(\dfrac{1}{n}\text{.}\)
Example 3.68.
For astronomers, the mass of a star is its most important property, but it is also the most difficult to measure directly. For many stars, their luminosity, or brightness, varies roughly as the fourth power of the mass.
-
Our Sun has luminosity \(4 \times 10^{26}\) watts and mass \(2 \times 10^{30}\) kilograms. Because the numbers involved are so large, astronomers often use these solar constants as units of measure:
The luminosity of the Sun is \(1\) solar luminosity, and its mass is \(1\) solar mass.
Write a power function for the luminosity, \(L\text{,}\) of a star in terms of its mass, \(M\text{,}\) using units of solar mass and solar luminosity.
The star Sirius is \(23\) times brighter than the Sun, so its luminosity is \(23\) solar luminosities. Estimate the mass of Sirius in units of solar mass.
Solution.
Because
\(L\) varies as the fourth power of
\(M\text{,}\) we have
\begin{equation*}
L = kM^4
\end{equation*}
Substituting the values of
\(L\) and
\(M\) for the Sun (namely,
\(L = 1\) and
\(M = 1\)), we find
\begin{equation*}
1 = k(1)^4
\end{equation*}
so
\(k = 1\) and
\(L = M^4\text{.}\)
We substitute the luminosity of Sirius,
\(L = 23\text{,}\) to get
\begin{equation*}
23 = M^4
\end{equation*}
To solve the equation for
\(M\text{,}\) we raise both sides to the
\(\dfrac{1}{4}\) power.
\begin{equation*}
\begin{aligned}[t]
(23)^{1/4} \amp = \left(M^4\right)^{1/4} \\
2.1899 \amp = M
\end{aligned}
\end{equation*}
The mass of Sirius is about
\(2.2\) solar masses, or about
\(2.2\) times the mass of the Sun.
Checkpoint 3.69. Practice 7.
A spherical fish tank in the lobby of the Atlantis Hotel holds about 905 cubic feet of water. What is the radius of the fish tank?
Solution.
Solve \(~\dfrac{4}{3}\pi r^3 = 905~ \) to find that the radius \(r\) is about \(6\) feet.
Subsection Power Functions
The basic functions \(y = \sqrt{x}\) and \(y = \sqrt[3]{x}\) are power functions of the form \(f (x) = x^{1/n}\text{,}\) and the graphs of all such functions have shapes similar to those two, depending on whether the index of the root is even or odd.
Figure (a) shows the graphs of
\begin{equation*}
y = x^{1/2}, ~~y = x^{1/4},~~ \text{and} ~~y = x^{1/6}
\end{equation*}
Figure (b) shows the graphs of
\begin{equation*}
y = x^{1/3},~~ y = x^{1/5},~~ \text{and}~~ y = x^{1/7}
\end{equation*}
We cannot take an even root of a negative number. (See
A Note on Roots of Negative Numbers at the end of this section.) Hence, if
\(n\) is even, the domain of
\(f (x) = x^{1/n}\) is restricted to nonnegative real numbers, but if
\(n\) is odd, the domain of
\(f (x) = x^{1/n}\) is the set of all real numbers.
We will also encounter power functions with negative exponents. For example, an animal’s heart rate is related to its size or mass, with smaller animals generally having faster heart rates. The heart rates of mammals are given approximately by the power function
\begin{equation*}
H(m) = km^{-1/4}
\end{equation*}
where \(m\) is the animal’s mass and \(k\) is a constant.
Example 3.70.
A typical human male weighs about \(70\) kilograms and has a resting heart rate of \(70\) beats per minute.
Find the constant of proportionality, \(k\text{,}\) and write a formula for \(H(m)\text{.}\)
-
Fill in the table with the heart rates of the mammals whose masses are given.
Animal |
Shrew |
Rabbit |
Cat |
Wolf |
Horse |
Polar bear |
Elephant |
Whale |
Mass (kg) |
\(0.004\) |
\(2\) |
\(4\) |
\(80\) |
\(300\) |
\(600\) |
\(5400\) |
\(70,000\) |
Heart rate |
\(\) |
\(\) |
\(\) |
\(\) |
\(\) |
\(\) |
\(\) |
\(\) |
Sketch a graph of \(H\) for masses up to \(6000\) kilograms.
Solution.
We substitute \(H = 70\) and \(m = 70\) into the equation; then solve for \(k\text{.}\)
\begin{equation*}
\begin{aligned}[t]
70 \amp = k \cdot 70^{-1/4} \\
k \amp= \frac{70}{70^{-1/4}} = 70^{5/4} \approx 202.5 \\
\end{aligned}
\end{equation*}
Thus, \(H(m) = 202.5m^{-1/4}\text{.}\)
-
We evaluate the function \(H\) for each of the masses given in the table.
Animal |
Shrew |
Rabbit |
Cat |
Wolf |
Horse |
Polar bear |
Elephant |
Whale |
Mass (kg) |
\(0.004\) |
\(2\) |
\(4\) |
\(80\) |
\(300\) |
\(600\) |
\(5400\) |
\(70,000\) |
Heart rate |
\(805\) |
\(170\) |
\(143\) |
\(68\) |
\(49\) |
\(41\) |
\(24\) |
\(12\) |
-
We plot the points in the table to obtain the graph shown below.
Many properties relating to the growth of plants and animals can be described by power functions of their mass. The study of the relationship between the growth rates of different parts of an organism, or of organisms of similar type, is called allometry. An equation of the form
\begin{equation*}
\text{variable} = k(\text{mass})^p
\end{equation*}
used to describe such a relationship is called an allometric equation.
Of course, power functions can be expressed using any of the notations we have discussed. For example, the function in
Example 3.70 can be written as
\begin{equation*}
H(m) = 202.5m^{-1/4}~~~ \text{ or }~~~ H(m) = 202.5m^{-0.25}~~~ \text{ or} ~~~ H(m) = \frac{202.5}{\sqrt[4]{m}}
\end{equation*}
Checkpoint 3.71. Practice 8.
-
Complete the table of values for the power function \(f (x) = x^{-1/2}\text{.}\)
\(x\) |
\(0.1\) |
\(0.25\) |
\(0.5\) |
\(1\) |
\(2\) |
\(f(x)\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(x\) |
\(4\) |
\(8\) |
\(10\) |
\(20\) |
\(200\) |
\(f(x)\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
Sketch the graph of \(y = f (x)\text{.}\)
Write the formula for \(f (x)\) with a decimal exponent, and with radical notation.
Solution.
-
\(x\) |
\(0.1\) |
\(0.25\) |
\(0.5\) |
\(1\) |
\(2\) |
\(f(x)\) |
\(3.2\) |
\(2\) |
\(1.4\) |
\(1\) |
\(0.71\) |
\(x\) |
\(4\) |
\(8\) |
\(10\) |
\(20\) |
\(200\) |
\(f(x)\) |
\(0.5\) |
\(0.35\) |
\(0.32\) |
\(0.22\) |
\(0.1\) |
\(f(x) = x^{-0.5}\text{,}\) \(f(x)=\dfrac{1}{\sqrt{x}} \)
Checkpoint 3.72. Pause and Reflect.
Describe the differences in the graphs of \(f(x)=x^{\frac{1}{n}}\) for \(n\) positive and negative, for \(x \gt 0\text{.}\)
Subsection Solving Radical Equations
A radical equation is one in which the variable appears under a square root or other radical. The radical may be denoted by a fractional exponent. For example, the equation
\begin{equation*}
5x^{1/3} = 32
\end{equation*}
is a radical equation because \(x^{1/3} = \sqrt[3]{x}\text{.}\) To solve the equation, we first isolate the power to get
\begin{equation*}
x^{1/3} = 6.4
\end{equation*}
Then we raise both sides of the equation to the reciprocal of \(\dfrac{1}{3}\text{,}\) or \(3\text{.}\)
\begin{equation*}
\begin{aligned}[t]
\left(x^{1/3}\right)^3 \amp = 6.4^3 \\
x \amp = 262.144
\end{aligned}
\end{equation*}
Example 3.73.
When a car brakes suddenly, its speed can be estimated from the length of the skid marks it leaves on the pavement. A formula for the car’s speed, in miles per hour, is \(v = f (d) = (24d)^{1/2}\text{,}\) where the length of the skid marks, \(d\text{,}\) is given in feet.
If a car leaves skid marks \(80\) feet long, how fast was the car traveling when the driver applied the brakes?
How far will a car skid if its driver applies the brakes while traveling \(80\) miles per hour?
Solution.
To find the velocity of the car, we evaluate the function for \(d = \alert{80}\text{.}\)
\begin{equation*}
\begin{aligned}[t]
v\amp= (24 \cdot \alert{80})^{1/2} \\
\amp = (1920)^{1/2} \approx 43.8178046\\
\end{aligned}
\end{equation*}
The car was traveling at approximately \(44\) miles per hour.
-
We would like to find the value of \(d\) when the value of \(v\) is known. We substitute \(v = \alert{80}\) into the formula and solve the equation
\begin{equation*}
\alert{80} = (24d)^{1/2} ~~ \blert{\text{ Solve for }d.}
\end{equation*}
Because \(d\) appears to the power \(\frac{1}{2}\text{,}\) we first square both sides of the equation to get
\begin{equation*}
\begin{aligned}[t]
80^2 \amp = \left((24d)^{1/2}\right)^2 \amp\amp \blert{\text{Square both sides.}}\\
6400 \amp = 24d \amp\amp \blert{\text{Divide by }24.}\\
266.\overline{6} \amp = d
\end{aligned}
\end{equation*}
You can check that this value for \(d\) works in the original equation. Thus, the car will skid approximately \(267\) feet. A graph of the function \(v = (24d)^{1/2}\) is shown below, along with the points corresponding to the values in parts (a) and (b).
Checkpoint 3.74. QuickCheck 4.
Which law of exponents do we use when solving \(24d^{\frac{1}{2}} = 80\text{?}\)
\(\displaystyle (x+y)^n=x^n+y^n\)
\(\displaystyle (x^n)^m=x^{nm}\)
\(\displaystyle (xy)^n=x^ny^n\)
\(\displaystyle x^n x^m=x^{n+m}\)
Note 3.75.
Thus, we can solve an equation where one side is an \(n\)th root of \(x\) by raising both sides of the equation to the \(n\)th power. We must be careful when raising both sides of an equation to an even power, since extraneous solutions may be introduced. However, because most applications of power functions deal with positive domains only, they do not usually involve extraneous solutions.
Checkpoint 3.76. Practice 9.
What would be the mass of an animal whose heart rate is \(120\) beats per minute?
Solution.
We solve \(~120=202.5m^{-1/4}~\) to find \(m\approx 8.1\) kg.
Subsection A Note on Roots of Negative Numbers
You already know that \(\sqrt{-9}\) is not a real number, because there is no real number whose square is \(-9\text{.}\) Similarly, \(\sqrt[4]{-16}\) is not a real number, because there is no real number \(r\) for which \(r^4 = -16\text{.}\) (Both of these radicals are complex numbers. Complex numbers are discussed in Chapter 7.) In general, we cannot find an even root (square root, fourth root, and so on) of a negative number.
On the other hand, every positive number has two even roots that are real numbers. For example, both \(3\) and \(-3\) are square roots of \(9\text{.}\) The symbol \(\sqrt{9}\) refers only to the positive, or principal root, of \(9\text{.}\) If we want to refer to the negative square root of \(9\text{,}\) we must write \(-\sqrt{9} = -3\text{.}\) Similarly, both \(2\) and \(-2\) are fourth roots of \(16\text{,}\) because \(2^4 = 16\) and \((-2)^4 = 16\text{.}\) However, the symbol \(\sqrt[4]{16}\) refers to the principal, or positive, fourth root only. Thus,
\begin{equation*}
\sqrt[4]{16} = 2 ~~ \text{ and } ~~ -\sqrt[4]{16} = -2
\end{equation*}
Things are simpler for odd roots (cube roots, fifth roots, and so on). Every real number, whether positive, negative, or zero, has exactly one real-valued odd root. For example,
\begin{equation*}
\sqrt[5]{32} = 2 ~~ \text{ and } ~~ \sqrt[5]{-32} = -2
\end{equation*}
Here is a summary of our discussion.
Roots of Real Numbers.
Every positive number has two real-valued roots, one positive and one negative, if the index is even.
A negative number has no real-valued root if the index is even.
Every real number, positive, negative, or zero, has exactly one real-valued root if the index is odd.
Example 3.77.
\(\sqrt[4]{-625}\) is not a real number.
\(\displaystyle - \sqrt[4]{625} = -5\)
\(\displaystyle \sqrt[5]{-1} = -1\)
\(\sqrt[4]{-1}\) is not a real number.
The same principles apply to powers with fractional exponents. Thus
\begin{equation*}
(-32)^{1/5} = -2
\end{equation*}
but \((-64)^{1/6}\) is not a real number. On the other hand,
\begin{equation*}
-64^{1/6} = -2
\end{equation*}
because the exponent \(1/6\) applies only to \(64\text{,}\) and the negative sign is applied after the root is computed.
Checkpoint 3.78. QuickCheck 5.
Which of the following is undefined for negative \(x\text{?}\)
\(\displaystyle x^{\frac{1}{4}}\)
\(\displaystyle x^{-4}\)
\(\displaystyle (-x)^{\frac{1}{4}}\)
\(\displaystyle (-x)^{-4}\)
Checkpoint 3.79. Practice 10.
Evaluate each power, if possible. Enter "DNE" if it is not possible to evaluate.
\(\displaystyle -81^{1/4}\)
\(\displaystyle (-81)^{1/4}\)
\(\displaystyle -64^{1/3}\)
\(\displaystyle (-64)^{1/3}\)
Solution.
\(\displaystyle -3\)
undefined
\(\displaystyle -4\)
\(\displaystyle -4\)
Checkpoint 3.80. Pause and Reflect.
What is the domain of the function \(f(x)=x^{\frac{1}{2n}}\text{,}\) and why?