| | |
Solve the following equations. Pay special attention to domain issues which may require you to check your final solutions, or extraneous solutions introduced by actions taken on both sides of an equation that are not reversible.
$\displaystyle{2x^{-2} + 3x^{-1} + 1 = 0}$
$\displaystyle{3 \left( \frac{5}{x-1} \right)^2 + \left( \frac{5}{x-1} \right) - 2 = 0}$
$\displaystyle{\frac{2x-1}{x^2+2x-8} - \frac{3}{2-x} = \frac{2}{x+4}}$
$\displaystyle{\log_{10} (2x+50) = 2}$
$\displaystyle{\log_2 x + \log_2 (x-2) = 3}$
$\displaystyle{1+ \sqrt{2x+1} = x}$
$\displaystyle{\log_3 (7-x) = \log_3 (1-x) + 1}$
$\displaystyle{\sqrt{x+1} - 3x = 1}$
$\displaystyle{3x(x-1)^{1/2} + 2(x-1)^{3/2} = 0}$
$\displaystyle{\frac{\log_3 16}{2\log_3 x} = 2}$
$\displaystyle{4x^{-2} - x^{-1} - 5 = 0}$
$\displaystyle{\left( \frac{x+1}{x} \right)^2 + 2 \left( \frac{x+1}{x} \right) - 3 = 0}$
$\displaystyle{\sqrt{x} - \sqrt{x-5} = 1}$
SOLUTIONS (WITH WORK) TO PROBLEM #1
$\displaystyle{x = -1, -2}$
$\displaystyle{x = -4, \frac{17}{2}}$
$\displaystyle{x = -5}$
$\displaystyle{x = 25}$
$\displaystyle{x = 4 \textrm{ only}}$
$\displaystyle{x = 4 \textrm{ only}}$
$\displaystyle{x = -2}$
$\displaystyle{x = 0}$
$\displaystyle{x = 1}$
$\displaystyle{x = 2}$
$\displaystyle{x = \frac{4}{5}, -1}$
$\displaystyle{x = -\frac{1}{4}}$
$\displaystyle{x = 9}$
Solve the following equations.
$\displaystyle{\log_{10} \frac{1}{x^2} = 2}$
$\displaystyle{\log_3 \sqrt{x^2 + 17} = 2}$
$\displaystyle{\log_2 2x + \log_2 (x + \textstyle{\frac{3}{2}}) = 1}$
$\displaystyle{\log_2 (\log_3 x) = 2}$
$\displaystyle{\log_4 x^2 = (\log_4 x)^2}$
$\displaystyle{2\log_2 (x+2) - \log_2 (x^2 - 4) = 3}$
$\displaystyle{\log_6 2x - \log_6 (x+1) = 0}$
$\displaystyle{\frac{\log_2 8^x}{\log_2 \textstyle{\frac{1}{4}}} = \frac{1}{2}}$
$\displaystyle{\log_9 \sqrt{10x+5} - \textstyle{\frac{1}{2}} = \log_9 \sqrt{x+1}}$
$(\ln x)^2 = \ln x^2$
$\log_2 (x-7) + \log_2 x = \ln e^3$
$e^{\ln 2x} - \ln e^{3x} = -3$
$x = \pm \frac{1}{10}$
$x = \pm 8$
$x = \frac{1}{2}$
$x = 81$
$x = 1,\ 16$
$x = \frac{18}{7}$
$x = 1$
$x = -\frac{1}{3}$
$x = 4$
$1,e^2$
$8$ only
$3$
Solve the following equations.
$\displaystyle{\frac{x-8}{3} + \frac{x-3}{2} = 0}$
$\displaystyle{\frac{x^2}{x-3} = \frac{9}{x-3}}$
$\displaystyle{\frac{2}{3x+6} + \frac{1}{x^2-4}=\frac{4}{x-2}}$
$\displaystyle{\sqrt{2x-5} - \sqrt{x-3} = 1}$
$\displaystyle{\sqrt{3x+1} = 4}$
$\displaystyle{5+\sqrt{x+7} = x}$
$\displaystyle{\sqrt{x-3} + \sqrt{x+5} = 4}$
Solve the following equations.
$\displaystyle{x=5}$
$\displaystyle{x=-3}$
$\displaystyle{x=\textstyle{-\frac{5}{2}}}$
$\displaystyle{x=3,7}$
$\displaystyle{x=5}$
$\displaystyle{x=9}$
$\displaystyle{x=4}$