4.3 Solving Equations Involving Root and Power Functions

Vanessa Coffelt

4.3 Solving Equations Involving Root and Power Functions

In this section, we set about solving equations and inequalities involving power functions. Our first example demonstrates the usual sorts of strategies to employ when solving equations.

Example 4.3.1

Example 4.3.1.1

Solve the following equations analytically and verify your answers using a graph.

$(7-x)^{\frac{3}{2}} = 8$

Solution:

Solve $(7-x)^{\frac{3}{2}} = 8$ for $x$ .

One way to proceed to solve $(7-x)^{\frac{3}{2}} = 8$ is to use Definition 4.3 to rewrite $(7-x)^{\frac{3}{2}}$ as either $(\sqrt{7-x})^3$ or $\sqrt{(7-x)^3}$ . We opt for the former given $8$ is a perfect cube:

$\begin{array}{rclr} (7-x)^{\frac{3}{2}} & = & 8 & \\ (\sqrt{7-x})^3 & = & 8 & \text{rewrite using Definition 4.3} \\ \sqrt[3]{(\sqrt{7-x})^3} & = & \sqrt[3]{8} & \text{extract cube roots} \\ \sqrt{7-x} & = & 2 & \sqrt[3]{u^3}= u$ \\ 7-x & = & 4 & \text{square both sides} \\ x & = & 3 & \\ \end{array}$

Thus, our solution is $x=3$ .

We verify our answer analytically by substituting $x=3$ into the original equation.

Geometrically, we are looking for where the graph of $f(x) = (7-x)^{\frac{3}{2}}$ intersects the graph of $g(x) = 8$ . Below are those graphs and we see the intersection point of both graphs is $(3,8)$ , thereby checking our solution $x = 3$ .

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Example 4.3.1.2

Solve the following equations analytically and verify your answers using a graph.

$(2t-1)^{\frac{2}{3}} -4 = 0$

Solution:

Solve $(2t-1)^{\frac{2}{3}} -4 = 0$ for $x$ .

Proceeding similarly to the above, to solve $(2t-1)^{\frac{2}{3}} -4 = 0$ , we rewrite $(2t-1)^{\frac{2}{3}}$ as $(\sqrt[3]{2t-1})^2$ and solve:

$\begin{array}{rclr} (2t-1)^{\frac{2}{3}} -4 & = & 0 & \\ (\sqrt[3]{2t-1})^2 - 4 & = & 0 & \text{rewrite using Definition 4.3} \\ (\sqrt[3]{2t-1})^2 & = & 4 & \text{isolate the variable term} \\ \sqrt{(\sqrt[3]{2t-1})^2 } & = & \sqrt{4} & \text{extract square roots} \\ |\sqrt[3]{2t-1}| & = & 2 & \sqrt{u^2} = |u| \\ \sqrt[3]{2t-1} & = & \pm 2 & \text{ for } c>0, \, |u| = c \text{ is equivalent to } u = \pm c. \end{array}$

From $\sqrt[3]{2t-1} = \pm 2$ we cube both sides of each equation.

$\begin{array}{rclcrcl} \sqrt[3]{2t-1} & = & -2 & \text{ and } & \sqrt[3]{2t-1} & = & 2 \\ 2t - 1 & = & -8 & \qquad & 2t-1 & = & 8 \\ t & = & - \frac{7}{2} & \qquad & t & = & \frac{9}{2} \end{array}$

Both of $t = -\frac{7}{2} = -3.5$ and $t= \frac{9}{2} = 4.5$ are solutions to the given equation.

In this case we are looking for where the graph of $f(t) = (2t-1)^{\frac{2}{3}} -4$ intersects the graph of $g(t) = 0$ – i.e., the $t$ -intercepts of the graph of $g$ . We find these are $(-3.5,0)$ and $(4.5,0)$ , as predicted.

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Example 4.3.1.3

Solve the following equations analytically and verify your answers using a graph.

$(x+3)^{0.5} = 2(7-x)^{0.5}+1$

Solution:

Solve $(x+3)^{0.5} = 2(7-x)^{0.5}+1$ for $x$ .

$0.5 = \frac{1}{2}$ , so we may rewrite $(x+3)^{0.5} = 2(7-x)^{0.5}+1$ as $(x+3)^{\frac{1}{2}} = 2(7-x)^{\frac{1}{2}}+1$ . Using Definition 4.3, we then have $\sqrt{x+3} = 2\sqrt{7-x} + 1$ . One of the square roots is already isolated, thus we can rid ourselves of it by squaring both sides.

$\begin{array}{rclr} \sqrt{x+3} & = & 2\sqrt{7-x} + 1 & \\ (\sqrt{x+3})^2 & = & (2\sqrt{7-x} + 1)^2 & \text{square both sides} \\ x+3 & = & (2 \sqrt{7-x})^2 + 2 (2 \sqrt{7-x})(1) + 1 & (\sqrt{u})^2 = u \text{ and } (a+b)^2 = a^2 + 2ab +b^2 \\ x+3 & = & 4(7-x) + 4\sqrt{7-x} + 1 & (ab)^2 = a^2b^2 \text{ and, again, } (\sqrt{u})^2 = u \\ x+3 & = & 28-4x+4\sqrt{7-x} + 1 & \\ 5x-26 & = & 4\sqrt{7-x} & \text{isolate } \sqrt{7-x} \\ \end{array}$

We square both sides again and get $(5x-26)^2 = (4\sqrt{7-x})^2$ .

$\begin{array}{rclr} 25x^2-260x+676 & = & 16(7-x) & \text{square both sides} \\ 25x^2-244x+564 & =& 0 & \text{set quadratic equal to zero} \\ (x-6)(25x-94) & = & 0 & \text{ factor} \end{array}$

We set each factor equal to zero, using the Zero Product Property to get

$\begin{array}{rclcrcl} x-6 & = & 0 & \text{ or } & 25x-94 & = & 0 \\ x & = & 6 & \qquad & x & = & \frac{94}{25} \end{array}$

When we go to check these answers, we verify $x=6$ is a solution, but $x = 3.76$ is not. Hence, $x=3.76$ is an `extraneous’ solution.^[1]

We graph both $f(x) = \sqrt{x+3}$ and $g(x) = 2\sqrt{7-x} + 1$ below and confirm there is only one intersection point, $(6,3)$ .

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Example 4.3.1.4

Solve the following equations analytically and verify your answers using a graph.

$2t^{\frac{2}{3}} + 5t^{\frac{1}{3}} = 3$

Solution:

Solve $2t^{\frac{2}{3}} + 5t^{\frac{1}{3}} = 3$ for $x$ .

While we could approach solving $2t^{\frac{2}{3}} + 5t^{\frac{1}{3}} = 3$ as the previous example, we would encounter cubing binomials^[2] which we would prefer to avoid. Instead, we take a step back and notice there are three terms here with the exponent on one term, $t^{\frac{2}{3}}$ exactly twice the exponent on another term, $t^{\frac{1}{3}}$ . We have ourselves a `quadratic in disguise.’^[3]

To help us see the forest for the trees, we let $u = t^{\frac{1}{3}}$ so that $u^2 = t^{\frac{2}{3}}$ . (Note that root here, $3$ , is odd, so we can use the properties of exponents stated in Theorem 4.3.) Hence,

$\begin{array}{rclr} 2t^{\frac{2}{3}} + 5t^{\frac{1}{3}} & = & 3 & \text{ in terms of } t \\ 2u^2 + 5u & = & 3 & \text{in terms of } u \\ 2u^2 + 5u - 3 & = & 0 & \\ (2u-1)(u+3) & = & 0 & \text{factoring} \end{array}$

We set each factor equal to zero, using the Zero Product Property

$\begin{array}{rclcrclr} 2u - 1 & = & 0 & \text{ or } & u+3 & = & 0 & \\[4pt] u & = & \frac{1}{2} & \qquad & u & = & -3 & \\[4pt] t^{\frac{1}{3}} & = & \frac{1}{2} & \text{ or } & t^{\frac{1}{3}} & = & -3 & \text{ replacing } u = t^{\frac{1}{3}} \\[4pt] t & = & \frac{1}{8} & \qquad & t & = & -27 & \text{cube both sides} \end{array}$

Both of $t = \frac{1}{8} = 0.125$ and $t = -27$ are solutions to our original equation.

Looking at the graphs of $f(t) = 2t^{\frac{2}{3}} + 5t^{\frac{1}{3}}$ and $g(t) = 3$ , we find two intersection points, $(-27,3)$ and $(0.125,3),$ as required.

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Example 4.3.1.5

Solve the following equations analytically and verify your answers using a graph.

$2(3x-1)^{-0.5} = 3x (3x-1)^{-1.5}$

Solution:

Solve $2(3x-1)^{-0.5} = 3x (3x-1)^{-1.5}$ for $x$ .

Next we are to solve $2(3x-1)^{-0.5} = 3x (3x-1)^{-1.5}$ which, when written without negative exponents is: $\frac{2}{(3x-1)^{0.5}} = \frac{3x}{(3x-1)^{1.5}}.$

The rational exponents here are $0.5 = \frac{1}{2}$ and $1.5 = \frac{3}{2}$ , both involve an even indexed root (the square root in this case!) which means $3x-1 \geq 0.$

Moreover, the $3x-1$ resides in the denominator meaning $3x - 1 \neq 0$ so our equation is really valid only for values of $x$ where $3x-1>0$ or $x > \frac{1}{3}.$

Hence, we clear denominators and can apply Theorem 4.3:

$\begin{array}{rclr} \dfrac{2}{(3x-1)^{0.5}} & = & \dfrac{3x}{(3x-1)^{1.5}} & \\ \left[ \dfrac{2}{(3x-1)^{0.5}} \right] \cdot (3x-1)^{1.5} & = & \left[ \dfrac{3x}{(3x-1)^{1.5}} \right ] \cdot (3x-1)^{1.5} & \\ 2 \cdot \dfrac{(3x-1)^{1.5}}{(3x-1)^{0.5}} & = & 3x & \\ 2 (3x-1)^{1.5-0.5} & = & 3x & \text{Theorem 4.3 applies as $3x-1 > 0$.} \\ 2 (3x-1)^{1} & = & 3x & \\ 6x-2 & = & 3x & \\ x & = & \frac{2}{3} & \end{array}$

Because $x = \frac{2}{3} > \frac{1}{3}$ , we keep it and, sure enough, it is a solution to our original equation.

Graphically we see $f(x)=2(3x-1)^{-0.5}$ intersects $g(x) = 3x (3x-1)^{-1.5}$ at $\left(\frac{2}{3}, 2\right).$

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Example 4.3.1.6

Solve the following equations analytically and verify your answers using a graph.

$6(9-t^2)^{\frac{1}{3}} = 4t^2 (9-t^2)^{-\frac{2}{3}}$

Solution:

Solve $6(9-t^2)^{\frac{1}{3}} = 4t^2 (9-t^2)^{-\frac{2}{3}}$ for $x$ .

Our last equation to solve is $6(9-t^2)^{\frac{1}{3}} = 4t^2 (9-t^2)^{-\frac{2}{3}}$ , which, when rewritten without negative exponents is: $6(9-t^2)^{\frac{1}{3}} = \frac{4t^2}{(9-t^2)^{\frac{2}{3}}}$ . Again, the root here ( $3$ ) is odd, so we can use the exponent properties listed in Theorem 4.3. We begin by clearing denominators:

$\begin{array}{rclr} 6(9-t^2)^{\frac{1}{3}} & = & \frac{4t^2}{(9-t^2)^{\frac{2}{3}}} & \\[10pt] 6(9-t^2)^{\frac{1}{3}} \cdot (9-t^2)^{\frac{2}{3}} & = & \left[ \frac{4t^2}{(9-t^2)^{\frac{2}{3}}} \right ] \cdot (9-t^2)^{\frac{2}{3}} & \\ 6(9-t^2)^{\frac{1}{3} + \frac{2}{3}}& = & 4t^2 & \text{Theorem 4.3 applies because the root here $3$ is odd.} \\ 6(9-t^2)^{1} & = & 4t^2 &\\ 54 - 6t^2 & = & 4t^2 & \\ 10t^2 & = & 54 & \\ t^2 & = & \frac{54}{10} & \\[6pt] t & = & \pm \sqrt{\frac{27}{5}} & \\[6pt] t & = & \pm 3 \sqrt{15}{5} \\[6pt] \end{array}$

So $t = \pm 3 \sqrt{15}{5} \approx \pm 2.324$ . While not the easiest to check analytically, both of these solutions do work in the original equation.

Graphing $f(t) = 6(9-t^2)^{\frac{1}{3}}$ and $g(t) = 4t^2 (9-t^2)^{-\frac{2}{3}}$ below, we see the graphs intersect when $t \approx \pm 2.324$ .

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Note that in Example 4.3.1, there are several ways to correctly solve each equation, and we endeavored to demonstrate a variety of methods. For example, for number 1, instead of converting $(7-x)^{\frac{3}{2}}$ to a radical equation, we could use Theorem 4.3. The root here ( $2$ ) is even, thus we know $7-x \geq 0$ or $x \leq 7$ . Hence we may apply exponent properties:

$\begin{array}{rclr} (7-x)^{\frac{3}{2}} & = & 8 & \\ \left[(7-x)^{\frac{3}{2}}\right]^{\frac{2}{3}} & = & 8^{\frac{2}{3}} & \text{raise both sides to the $\frac{2}{3}$ power} \\ (7-x)^{\frac{3}{2} \cdot \frac{2}{3}} & = & 4 & \text{Theorem 4.3} \\ (7-x)^{1} & = & 4 \\ \end{array}$

from which we get $x = 3$ . If we try this same approach to solve number 2, however, we encounter difficulty. From $(2t-1)^{\frac{2}{3}} -4 = 0$ , we get $(2t-1)^{\frac{2}{3}} =4$ .

$\begin{array}{rclr} (2t-1)^{\frac{2}{3}} & = & 4 & \\ \left[(2t-1)^{\frac{2}{3}} \right]^{\frac{3}{2}} & = & 4^{\frac{3}{2}}& \text{raise both sides to the $\frac{3}{2}$ power} \\ \end{array}$

Given the root here ( $3$ ) is odd, we have no restriction on $2t-1$ but the exponent $\frac{3}{2}$ has an even denominator. Hence, Theorem 4.3 does not apply. That is,

$\begin{array}{rcl} \left[(2t-1)^{\frac{2}{3}} \right]^{\frac{3}{2}} & \neq & (2t-1)^{\frac{2}{3} \cdot \frac{3}{2}} \\ & = & (2t-1)^{1} \\ & = & (2t-1) \end{array}$

Note that if we weren’t careful, we’d have $2t-1 = 4^{\frac{3}{2}} = 8$ which gives $t= \frac{9}{2} = 4.5$ only. We’d have missed the solution $t = -3.5$ . Truth be told, you can simplify $\left[(2t-1)^{\frac{2}{3}} \right]^{\frac{3}{2}}$ – just not using Theorem 4.3. We leave it as an exercise to show $\left[(2t-1)^{\frac{2}{3}} \right]^{\frac{3}{2}} = |2t-1|$ and, more generally, $\left(x^{\frac{2}{3}}\right)^{\frac{3}{2}} = |x|$ .

Our next example is an application of the Cobb Douglas production model of an economy. The Cobb-Douglas model states that the yearly total dollar value of the production output in an economy is a function of two variables: labor (the total number of hours worked in a year) and capital (the total dollar value of the physical goods required for manufacturing.) The equation relating the production output level $P$ , labor $L$ and capital $K$ takes the form $P = a L^{b} K^{1-b}$ where $0 < b < 1$ ; that is, the production level varies jointly with some power of the labor and capital.

Example 4.3.2

Example 4.3.2.1

In their original paper A Theory of Production^[4] Cobb and Douglas modeled the output of the US Economy (using 1899 as a baseline) using the formula

$P = 1.01 L^{0.75} K^{0.25}$

where $P$ , $L$ , and $K$ were percentages of the 1899 figures for total production, labor, and capital, respectively.

For 1910, the recorded labor and capital figures for the US Economy are 144% and 208% of the 1899 figures, respectively. Compute $P$ using these figures and interpret your answer.

Solution:

For 1910, the recorded labor and capital figures for the US Economy are 144% and 208% of the 1899 figures, respectively. Compute $P$ using these figures and interpret your answer.

In this case,

$\begin{array}{rcl} P & = & 1.01 L^{0.75} K^{0.25} \\ & = & 1.01 (144)^{0.75} (208)^{0.25} \\ & \approx & 159 \end{array}$

which means the dollar value of the total US Production in 1920 was approximately 159% of what it was in 1899.^[5]

Example 4.3.2.2

In their original paper A Theory of Production^[6] Cobb and Douglas modeled the output of the US Economy (using 1899 as a baseline) using the formula

$P = 1.01 L^{0.75} K^{0.25}$

where $P$ , $L$ , and $K$ were percentages of the 1899 figures for total production, labor, and capital, respectively.

The recorded production value figure for 1920 is 231% of the 1899 figure. Use this to write $K$ as a function of $L$ , $K = f(L)$ . Compute and interpret $f(193)$ .

Solution:

The recorded production value figure for 1920 is $231 \%$ of the 1899 figure. Use this to write $K$ as a function of $L$ , $K = f(L)$ . Compute and interpret $f(193)$ .

We are given $P = 231 = 1.01L^{0.75} K^{0.25}$ , so to write $K$ as a function of $L$ , we need to solve this equation for $K$ . $L$ and $K$ are positive by definition, so we can employ properties of exponents:

$\begin{array}{rclr} 231 & = & 1.01 L^{0.75} K^{0.25} & \\[5pt] \dfrac{231}{1.01 L^{0.75}} & = & \dfrac{1.01 L^{0.75} K^{0.25}}{1.01 L^{0.75}} & \text{$L>0$, hence $L^{0.75} \neq 0$.}\\[8pt] K^{0.25} & = & 228.\overline{7128}L^{-0.75} & \text{rewrite} \\[5pt] \left(K^{0.25}\right)^{\frac{1}{0.25}} & = & \left(228.\overline{7128}L^{-0.75}\right)^{\frac{1}{0.25}} & \\[5pt] K^{\frac{0.25}{0.25}} & = & (228.\overline{7128})^{\frac{1}{0.25}} L^{-\frac{0.75}{0.25}} & \text{Theorem 4.3} \\[5pt] K & = & (228.\overline{7128})^{4} L^{-3} & \text{simplify} \\ \end{array}$

Hence, $K = f(L) = (228.\overline{7128})^{4} L^{-3}$ where $L>0$ .

We find

$\begin{array}{rcl} f(193) & = & (228.\overline{7128})^{4} (193)^{-3} \\ & \approx & 381 \end{array}$

meaning that in order to maintain a production level of 231% of 1889 with a labor level at 193% of 1889, the required capital is $381\%$ that of 1889.^[7]

Example 4.3.2.3

In their original paper A Theory of Production^[8] Cobb and Douglas modeled the output of the US Economy (using 1899 as a baseline) using the formula

$P = 1.01 L^{0.75} K^{0.25}$

where $P$ , $L$ , and $K$ were percentages of the 1899 figures for total production, labor, and capital, respectively.

Graph $K = f(L)$ and interpret the behavior as $L \rightarrow 0^{+}$ and $L \rightarrow \infty$ .

Solution:

Graph $K = f(L)$ and interpret the behavior as $L \rightarrow 0^{+}$ and $L \rightarrow \infty .$

The function $f(L)$ is a Laurent Monomial (see Section 3.2) with $n = 3$ and $a = (228.\overline{7128})^{4}$ .

As such, as $L \rightarrow 0^{+}$ , $f(L) \rightarrow \infty$ . This means that in order to maintain the given production level, as the available labor diminishes, the capital requirement become unbounded.

As $L \rightarrow \infty$ , we have $f(L) \rightarrow 0$ meaning that as the available labor increases, the need for capital diminishes.

The graph of $f$ is called an `isoquant’ – meaning `same quantity.’ In this context, the graph displays all combinations of labor and capital, $(L,K)$ which result in the same production level, in this case, $231\%$ of what was produced in 1889.

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4.3.1 Section Exercises

In Exercises 1 – 14, solve the equation.

$x+1 = (3x+7)^{\frac{1}{2}}$
$2x+1 = (3-3x)^{\frac{1}{2}}$
$t + (3t+10)^{0.5} = -2$
$3t+(6-9t)^{0.5}=2$
$x^{-1.5} = 8$
$2x - 1 = (x + 1)^{-0.5}$
$t^{\frac{2}{3}} = 4$
$(t - 2)^{\frac{1}{2}} + (t - 5)^{\frac{1}{2}} = 3$
$(2x+1)^{\frac{1}{2}} = 3 + (4-x)^{\frac{1}{2}}$
$5 - (4-2x)^{\frac{2}{3}} = 1$
$2t^{\frac{2}{3}} = 6 - t^{\frac{1}{3}}$
$2t^{\frac{1}{3}} = 1-3t^{\frac{2}{3}}$
$2x^{1.5} = 15x^{0.75} + 8$
$35x^{-0.75} = x^{-1.5} +216$
The Cobb-Douglas production model^[9] for the country of Sasquatchia is $P = 1.25L^{0.4}K^{0.6}$ . Here, $P$ represents the country’s production (measured in thousands of Bigfoot Bullion), $L$ represents the total labor (measured in thousands of hours) and $K$ represents the total investment in capital (measured in Bigfoot Bullion.) Let $P = 300$ and solve for $K$ as a function of $L$ . If $L = 100$ , what is $K$ ? Interpret each of the quantities in this case.

Section 4.3 Exercise Answers can be found in the Appendix … Coming soon

We invite the reader to see at which point in our machinations $x=3.76$ does check. Knowing a solution is extraneous is one thing; understanding how it came about is another. ↵
that is, expanding things like $(a+b)^3$ . ↵
See Section 0.5.5. ↵
available here. ↵
This answer is remarkably accurate. Note: all the dollar values here are recorded in `1880' dollars, per the source article. ↵
available here. ↵
The actual recorded figure is 407. ↵
available here. ↵
See Example 4.3.2 for more details on these sorts of models. ↵

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Example 4.3.1

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Example 4.3.2

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