In Exercises powereqineqexfirsta - powereqineqexlasta, solve the equation or inequality.
\(x+1 = (3x+7)^{\frac {1}{2}}\)

\(x=\answer {3}\)

\(2x+1 = (3-3x)^{\frac {1}{2}}\)

\(x=\answer {\frac {1}{4}}\)

\(t + (3t+10)^{0.5} = -2\)

\(t=\answer {-3}\)

\(3t+(6-9t)^{0.5}=2\)
\(t=-\frac {2}{3}\) \(t=-\frac {1}{3}\) \(x=0\) \(t=\frac {1}{3}\) \(t=\frac {2}{3}\) \(x=3\)
\(x^{-1.5} = 8\)

\(x=\answer {\frac {1}{4}}\)

\(2x - 1 = (x + 1)^{-0.5}\)

\(x = \frac {\sqrt {3}}{2}\)
\(t^{\frac {2}{3}} = 4\)

\(t = \pm 8\)
\((t - 2)^{\frac {1}{2}} + (t - 5)^{\frac {1}{2}} = 3\)

\(t = \answer {6}\)

\((2x+1)^{\frac {1}{2}} = 3 + (4-x)^{\frac {1}{2}}\)

\(x=\answer {4}\)

\(5 - (4-2x)^{\frac {2}{3}} = 1\)

\(x=-2, 6\)
\(2t^{\frac {2}{3}} = 6 - t^{\frac {1}{3}}\)

\(t=-8, \frac {27}{8}\)
\(2t^{\frac {1}{3}} = 1-3t^{\frac {2}{3}} \)

\(t=-1, \frac {1}{27}\)
\(2x^{1.5} = 15x^{0.75} + 8\)

\(x=16, \frac {1}{16}\)
\(35x^{-0.75} = x^{-1.5} +216\)

\(x = \frac {1}{81}, \frac {1}{16}\)
\(10-\sqrt {t-2} \leq 11\)

\([2, \infty )\)
\(t^{\frac {2}{3}} \leq 4\)

\([-8,8]\)
\(\sqrt [3]{x} \leq x\)

\([-1, 0] \cup [1, \infty )\)
\((2-3x)^{\frac {1}{3}} > 3x\)

\(\left (-\infty , \frac {1}{3} \right )\)
\((t^2-1)^{-\frac {1}{2}} \geq 2\)

\(\left [ -\frac {\sqrt {5}}{2}, -1\right ) \cup \left (1, \frac {\sqrt {5}}{2}\right ]\)
\((t^2-1)^{-\frac {1}{3}} \leq 2\)

\(\left (-\infty , -\frac {3\sqrt {2}}{4} \right ] \cup (-1,1) \cup \left [ \frac {3\sqrt {2}}{4}, \infty \right )\)
\(3(x-1)^{\frac {1}{3}} +x (x-1)^{-\frac {2}{3}} \geq 0\)

\(\left [ \frac {3}{4}, 1\right ) \cup (1, \infty )\)
\(3(x-1)^{\frac {2}{3}} +2x (x-1)^{-\frac {1}{3}} \geq 0\)

\(\left ( -\infty , \frac {3}{5} \right ] \cup (1, \infty )\)
\(2 (t-2)^{-\frac {1}{3}} -\frac {2}{3} t(t-2)^{-\frac {4}{3}} \leq 0\)

\((-\infty , 2) \cup (2,3]\)
\(-\frac {4}{3} (t-2)^{-\frac {4}{3}} + \frac {8}{9} t (t-2)^{-\frac {7}{3}} \geq 0\)

\((2,6]\)
\(2x^{-\frac {1}{3}}(x-3)^{\frac {1}{3}} + x^{\frac {2}{3}} (x-3)^{-\frac {2}{3}} \geq 0\)

\((-\infty , 0) \cup [2,3) \cup (3, \infty )\)
\(\sqrt [3]{x^{3} + 3x^{2} - 6x - 8} > x + 1\)

\((-\infty , -1)\)
\(4(7-t)^{0.75} - 3t(7-t)^{-0.25} \leq 0\)

\([4,7)\)
\(4t^{0.75}(t - 3)^{-\frac {2}{3}} +9t^{-0.25}(t - 3)^{\frac {1}{3}} < 0\)

\(\left (0, \frac {27}{13} \right )\)
\(x^{-\frac {1}{3}} (x-3)^{-\frac {2}{3}} - x^{-\frac {4}{3}} (x-3)^{-\frac {5}{3}} (x^2-3x+2) \geq 0\)

\((-\infty , 0) \cup (0,3)\)
\(\frac {2}{3}(t + 4)^{\frac {3}{5}}(t - 2)^{-\frac {1}{3}} + \frac {3}{5}(t + 4)^{-\frac {2}{5}}(t - 2)^{\frac {2}{3}} \geq 0\)

\((-\infty , -4) \cup \left (-4, -\frac {22}{19}\right ] \cup (2, \infty )\)
The Cobb-Douglas production model 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.

    \(K=f(L) = (240)^{ \frac {5}{3}} L^{- \frac {2}{3}}\). \(f(100) \approx 430.2148\). This means in order for the production level of Sasquatchia to reach 300,000 Bigfoot Bullion with a labor investment of 100,000 hours, the country needs to invest approximately 430 Bigfoot Bullion into capital.

  • Graph your answer to KintermsofLCobbexercise using a graphing utility. What information does an ordered pair \((L, K)\) on this graph represent?

    If a point \((L,K)\) is on the graph of this function, it means a combination of \(L\) thousand hours of labor with an investment of \(K\) Bigfoot Bullion into the Sasquatian Economy will result in a production level of 300,000 Bigfoot Bullion.
    PIC