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For each relation resulting from the given equation and any additional restrictions provided, solve for the indicated variable. Then graph the relation:
$\displaystyle{x^2 + y^2 = 16, \ \ y \gt 0}$; solve for $y$
$\displaystyle{x^2 + y^2 = 9, \ \ y \le 0}$; solve for $y$
$\displaystyle{x^2 + y^2 = 25, \ \ x \lt 0}$; solve for $x$
$\displaystyle{x^2 + y^2 = 7, \ \ x \ge 0}$; solve for $x$
$\displaystyle{x=y^2 + 1, \ \ x \lt 0}$; solve for $y$
$\displaystyle{x=-2y^2, \ \ x \ge 0};$ solve for $y$
$\displaystyle{x^2 + y^2 = 0}$ solve for $y$
$y = \sqrt{16-x^2}$ with $x \neq \pm4$
$y = -\sqrt{9-x^2}$
$x = -\sqrt{25-x^2}$ with $y \neq \pm \sqrt{7}$
$x = \sqrt{7-y^2}$
no solution
$y = 0$
$y = 0$
Use an online graphing tool (like the one at www.desmos.com/calculator to graph the following relation $$\log_x y^2 + x = y$$ Based on what results, what appears to be the minimum number of functions needed to fully describe this relation? (Hint: recall that for a relation whose graph was a circle, we needed two functions)
$3$
Solve the following using Lambert's W function:
$xe^{2x} = 3$
$x \cdot 3^x = 9$
$x + \ln x = \ln 2$
$2^x = 10x$
$x^2 + 8\ln x = 0$
$x^2 = 2^x$
$\displaystyle{\begin{array}[t]{rcl} xe^{2x} &=& 3\\ 2x e^{2x} &=& 6\\ 2x &=& W(6)\\ x &=& \frac{1}{2} W(6) \end{array}}$
$\displaystyle{\begin{array}[t]{rcl} x \cdot 3^x &=& 9\\ x \cdot e^{\ln 3^x} &=& 9\\ x \cdot e^{x \ln 3} &=& 9\\ (x \ln 3) \cdot e^{x \ln 3} &=& 9 \ln 3\\ x \ln 3 &=& W(9 \ln 3)\\ x &=& \frac{1}{\ln 3} W(9 \ln 3) \end{array}}$
$\displaystyle{\begin{array}[t]{rcl} x + \ln x &=& \ln 2\\ e^{x + \ln x} &=& 2\\ e^{\ln x} \cdot e^x &=& 2\\ x \cdot e^x &=& 2\\ x &=& W(2) \end{array}}$
$\displaystyle{\begin{array}[t]{rcl} 2^x &=& 10x\\ 10x \cdot 2^{-x} &=& 1\\ 10x \cdot e^{-x \ln 2} &=& 1\\ (-x \ln 2) \cdot e^{-x \ln 2} &=& -\frac{1}{10} \ln 2\\ -x \ln 2 &=& W_0(-\frac{1}{10} \ln 2) \quad \textit{or} \quad W_1(-\frac{1}{10} \ln 2)\\ x &=& -\frac{1}{\ln 2} W_0(-\frac{1}{10} \ln 2) \quad \textit{or} \quad -\frac{1}{\ln 2}W_1(-\frac{1}{10} \ln 2)\\ \end{array}}$
$\displaystyle{\begin{array}[t]{rcl} x^2 + 8\ln x &=& 0\\ 8\ln x &=& -x^2\\ (8 \ln x) \cdot x^{-2} &=& -1\\ (8 \ln x) \cdot e^{-2\ln x} &=& -1\\ (-2\ln x) \cdot e^{-2\ln x} &=& \frac{1}{4}\\ -2\ln x &=& W(\frac{1}{4})\\ \ln x &=& -\frac{1}{2} W(\frac{1}{4})\\ x &=& e^{-\frac{1}{2} W(\frac{1}{4})} \end{array}}$
Notice that $x^2 = 2^x$ implies $\ln x^2 = \ln 2^x$. Now we consider two separate cases: if $x \gt 0$, we can then say $2\ln x = x \ln 2$. If $x \lt 0$, we have instead $2\ln (-x) = x\ln 2$. Both of these can be solved with Lambert's $W$.
The first yields two solutions: $$x = W(-\frac{1}{2} \ln 2) \quad \textit{or} \quad 4$$ The second yields one more solution: $$x = -e^{-W(\frac{1}{2} \ln 2)}$$