Question

calcule a soma de
$s = log_{100}0.1 + log_{25} \sqrt[3]{5} - log_{ \sqrt{2} }2$

Answer 1

$log_{100}(0.1) + log_{25}( \sqrt[3]{5} ) - log_{ \sqrt{2} }(2) = \\ \frac{ log(0.1) }{ log(100) } + \frac{ log_{5}( \sqrt[3]{5} ) }{ log_{5}(25) } - \frac{ log_{2}(2) }{ log_{2}( \sqrt{2} ) } = \\ \frac{ log( {10}^{ - 1} ) }{ log( {10}^{2} ) } + \frac{ log_{5 }( {5}^{ \frac{1}{3} } ) }{ log_{5}( {5}^{2} ) } - \frac{ log_{2}( {2}^{1} ) }{ log_{2}( {2}^{ \frac{1}{2} } ) } = \\ \frac{ - 1 log(10) }{2 log(10) } + \frac{ \frac{1}{3} log_{5}(5) }{2 log_{5}(5) } - \frac{1 log_{2}(2) }{ \frac{1}{2} log_{2}(2) } = \\ \frac{ - 1}{2} + \frac{ \frac{1}{3} }{2} - \frac{1}{ \frac{1}{2} } = \\ - \frac{1}{2} + \frac{1}{6} - 2 = - \frac{3 }{6} + \frac{1}{6} - \frac{12}{6} = \\ \frac{ - 3 + 1 - 12}{6} = \frac{ - 14}{6} = - \frac{7}{3} \\ =\boxed{ - 2. \overline3}$