| | |
Solve the following equations using the "socks and shoes" principle:
$\displaystyle{3x-7=0}$
$\displaystyle{\frac{5x^3+2}{7}=6}$
$\displaystyle{4 \left( \frac{\sqrt[3]{x+1}}{7} + 2 \right)^5 = \frac{1}{8}}$
$\displaystyle{\frac{1}{3-\frac{1}{x}}=\frac{4}{11}}$
$\displaystyle{(\sqrt[3]{x}+5)^{-1}=2}$
$\displaystyle{2x^4-8=0}$
$\displaystyle{\log_6 (x^2 + 11) = 2}$
$\displaystyle{2^{x^3 - 7} = 5}$
$\displaystyle{ \begin{array}[t]{rcl} 3x - 7 &=& 0\\ 3x &=& 7\\ x &=& \frac{7}{3} \end{array}}$
$\displaystyle{ \begin{array}[t]{rcl} \displaystyle{\frac{5x^3+2}{7}} &=& 6\\\\ 5x^3+2 &=& 42\\ 5x^3 &=& 40\\ x^3 &=& 8\\ x &=& 2 \end{array}}$
$\displaystyle{ \begin{array}[t]{rcl} \displaystyle{4 \left( \frac{\sqrt[3]{x+1}}{7} + 2 \right)^5} &=& \displaystyle{\frac{1}{8}}\\\\ \displaystyle{\left( \frac{\sqrt[3]{x+1}}{7} + 2 \right)^5} &=& \displaystyle{\frac{1}{32}}\\\\ \displaystyle{\frac{\sqrt[3]{x+1}}{7} + 2} &=& \displaystyle{\frac{1}{2}}\\\\ \displaystyle{\frac{\sqrt[3]{x+1}}{7}} &=& \displaystyle{-\frac{3}{2}}\\\\ \displaystyle{\sqrt[3]{x+1}} &=& -\frac{21}{2}\\\\ \displaystyle{x+1} &=& -\frac{9261}{8}\\\\ x &=& -\frac{9269}{8} \end{array}}$
$\displaystyle{ \begin{array}[t]{rcl} \displaystyle{\frac{1}{3-\frac{1}{x}}} &=& \displaystyle{\frac{4}{11}}\\\\ \displaystyle{3-\frac{1}{x}} &=& \cfrac{11}{4}\\\\ \displaystyle{-\frac{1}{x}} &=& -\cfrac{1}{4}\\\\ \displaystyle{\frac{1}{x}} &=& \cfrac{1}{4}\\\\ x &=& 4 \end{array}}$
$\displaystyle{ \begin{array}[t]{rcl} (\sqrt[3]{x}+5)^{-1} &=& 2\\ \sqrt[3]{x}+5 &=& \frac{1}{2}\\ \sqrt[3]{x} &=& -\frac{9}{2}\\ x &=& -\frac{729}{8} \end{array}}$
$\displaystyle{ \begin{array}[t]{rcl} 2x^4-8 &=& 0\\ 2x^4 &=& 8\\ x^4 &=& 4\\ x &=& \pm 4^{1/4} = \pm (2^2)^{1/4} = \pm 2^{1/2} = \pm \sqrt{2} \end{array}}$
$\displaystyle{ \begin{array}[t]{rcl} \log_6 (x^2 + 11) &=& 2\\ x^2 + 11 &=& 36\\ x^2 &=& 25\\ x &=& \pm 5 \end{array}}$
$\displaystyle{x = \sqrt[3]{7 + \log_2 5}}$
Find a way to proceed to create an equation (or perhaps, multiple equations) containing only a single occurrence of a variable each -- making them solvable with "socks and shoes". Then, solve the equation(s). (Hint: try different things -- among them: expanding products; collecting like terms; factoring and using the zero-product property; substituting a $u$ when seeing the square of an expression, etc.)
$\displaystyle{x(x^2+3)-x=2x-8}$
$\displaystyle{z = 36z^3}$
$\displaystyle{2x^3 + x^2 - 2x - 1 = 0}$
$\displaystyle{a^{2/3} - 3a^{1/3} - 10 = 0}$
$\displaystyle{(a^2 - 1)^2 - (a^2 - 1) - 2 = 0}$
$\displaystyle{(6x^3 + x^2 - 35x)(49 - x^4) = 0}$
$\displaystyle{\log_{10} (|x| + 1) - \log_{10} (|x|+2) = 0}$
$\displaystyle{5\sqrt[3]{x^2} - 4\sqrt[3]{x} = 1}$
$\displaystyle{x^3 + 2x^2 - 3ax = 6a}$, where $a$ is a constant
$\displaystyle{x = -2}$
$\displaystyle{z = 0, \pm \frac{1}{6}}$
$\displaystyle{x = \pm 1, - \frac{1}{2}}$
$\displaystyle{a = 125, -8}$
$\displaystyle{a = \pm \sqrt{3},0}$
$\displaystyle{x = \pm \sqrt{7^{\phantom{1}}}, 0, -\frac{5}{2}, \frac{7}{3}}$
this equation has no solution
$\displaystyle{x = 1, -\frac{1}{125}}$
$\displaystyle{x = -2, \pm \sqrt{3a}}$
Graph the following functions after thinking about the simpler functions that can be composed together (and the order in which they are composed) to form them. Then state the implicit domain and image/range of each.
$f(x) = |x+2| - 3$
$g(x) = (x-3)^2 + 4$
$h(x) = \log_2 (-x)$
$f(x) = 2^{-2x} + 1$
$f(x) = (x+5)^3-2$
$g(x) = 2x + 3$
$h(x) = -x^3$ (use a reflection across the $x$-axis)
$f(x) = (-x)^3$ (use a reflection across the $y$-axis)
$g(x) = -3|1-2x|+5$
$f(x) = -3(2-x)^2+1$
$f(x) = \cfrac{-1}{x+2}$
$f(x) = -\sqrt{x+4}+3$
$f(x) = 3^{-x} + 2 $
$f(x) = \log_3 |x|$
Solve for the unknown
$\ln x = 3$
$\ln \sqrt{e} = x$
$\ln e^{2x} = -\frac{1}{2}$
$\ln e^4 = x^2$
$e^{-\frac{1}{2} \ln x} = 4$
$\ln x^3 = \frac{1}{e}$
$\ln x = \ln 1 + \ln 2 + \ln 3 + \ln 4$
$e^3$
$\frac{1}{2}$
$-\frac{1}{4}$
$\pm 2$
$\frac{1}{16}$
$e^{\frac{1}{3e}}$
$24$
Solve for $x$:
$e^{x^2 - 1} = 0$
$e^{2x+1} = 7$
$e^{-\ln x} = x$
$e^{-\ln x} = 5$
$2 \ln x = 1$
$\ln x = -1$
$2^{\ln x} = 4$
$e^{\frac{1}{2} \ln(x+1)} = 3$
$\ln x = e$
$\ln \sqrt{e^x}=-3$
$x=\pm 1$
$x = \frac{\ln 7 - 1}{2}$
$x = 1$
$x = \frac{1}{5}$
$\sqrt{e}$
$\frac{1}{e}$
$e^2$
$8$
$e^e$
$-6$