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Find the values of the following:
$\ln \sqrt{e}$
$\ln \sqrt[3]{e^2}$
$\ln (e^2 \cdot e^3)$
$\ln (e^2)^3$
$\ln \frac{1}{\sqrt[3]{e^2}}$
$\ln \left(\frac{e^{3/2}}{e^2 \sqrt{e}} \right)$
$\ln \frac{\sqrt{e^3}}{e}$
$e^{-\ln 3}$
$e^{\frac{1}{2}\ln \frac{1}{16} - \frac{2}{3} \ln 27} - \ln e^{\frac{5}{4}}$
$\frac{1}{2}$
$\frac{2}{3}$
$5$
$6$
$-\frac{2}{3}$
$-1$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{-11}{9}$
If the given statement is in exponential form, write it in logarithmic form. If instead it is in logarithmic form, write it in exponential form:
$e^y = 3$
$e^5 = x$
$\ln x = 3$
$\ln 3x = -2$
$\ln e^2 = 2$
$\ln 3 = y$
$\ln x = 5$
$e^3 = x$
$e^{-2} = 3x$
$e^2 = e^2$
Simplify and write as one logarithm:
$3\ln 5 - \frac{1}{2}\ln 4 + \ln 8$
$\ln(x^3+8)-\ln(x^2-2x+4)-2\ln(x+2)$
$\ln 500$
$-\ln (x+2)$
Solve for the unknown
$\ln x = 3$
$\ln \sqrt{e} = x$
$\ln e^{2x} = -\frac{1}{2}$
$\ln e^4 = x^2$
$(\ln x)^2 = \ln x^2$
$e^{-\frac{1}{2} \ln x} = 4$
$\ln x^3 = \frac{1}{e}$
$\ln x = \ln 1 + \ln 2 + \ln 3 + \ln 4$
$\log_2 (x-7) + \log_2 x = \ln e^3$
$e^{\ln 2x} - \ln e^{3x} = -3$
$e^3$
$\frac{1}{2}$
$-\frac{1}{4}$
$\pm 2$
$1,e^2$
$\frac{1}{16}$
$e^{\frac{1}{3e}}$
$24$
$8$ only
$3$
Solve for $x$:
$e^{x^2 - 1} = 0$
$e^{2x+1} = 7$
$e^{-\ln x} = x$
$e^{-\ln x} = 5$
$2 \ln x = 1$
$\ln x = -1$
$2^{\ln x} = 4$
$e^{\frac{1}{2} \ln(x+1)} = 3$
$\ln x = e$
$\ln \sqrt{e^x}=-3$
$x=\pm 1$
$x = \frac{\ln 7 - 1}{2}$
$x = 1$
$x = \frac{1}{5}$
$\sqrt{e}$
$\frac{1}{e}$
$e^2$
$8$
$e^e$
$-6$