By comparing areas, show that

$\displaystyle \frac{1}{3}<\ln 1.5 < \frac{5}{12}$

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- Jun 15th 2009, 03:34 AMRobbBy comparing areas find the value of a log
By comparing areas, show that

$\displaystyle \frac{1}{3}<\ln 1.5 < \frac{5}{12}$ - Jun 15th 2009, 04:01 AMalexmahone
$\displaystyle ln 1.5=ln \frac{3}{2}=ln 3-ln 2=\int \frac{1}{x}$ (from 2 to 3)