Question

Здравствуйте! Нужна срочная помощь!

Answer 1

Ответ:

$log_{ ab}(a) = 4 \\ \frac{1}{ log_{a}(ab) } = 4 \\ \frac{1}{ log_{a}(a) + log_{a}(b) } = 4 \\ \frac{1}{1 + log_{a}(b) } = 4 \\ 1 + log_{a}(b) = \frac{1}{4} \\ log_{a}(b) = - \frac{3}{4} \\ \frac{1}{ log_{b}(a) } = - \frac{3}{4} \\ log_{b}(a) = - \frac{4}{3}$

$\\ log_{ab}( \frac{ \sqrt[3]{a} }{ \sqrt{b} } ) = log_{ab}( \sqrt[3]{a} ) - log_{ab}( \sqrt{b} ) = \\ = \frac{1}{3} log_{ab}(a) - \frac{1}{2} log_{ab}(b) = \\ = \frac{1}{3 log_{a}(ab) } - \frac{1}{2 log_{b}(ab) } = \\ = \frac{1}{3( log_{a}(a) + log_{a}(b) )} - \frac{1}{2( log_{b}(a) + log_{b}(b) )} = \\ = \frac{1}{3(1 + log_{a}(b) )} - \frac{1}{2(1 + log_{b}(a) )} \\ \\ \frac{1}{3(1 - \frac{3}{4}) } - \frac{1}{2(1 - \frac{4}{3} )} = \frac{1}{3 \times \frac{1}{4} } - \frac{1}{2 \times ( - \frac{1}{3}) } = \\ = \frac{4}{3} + \frac{3}{2} = \frac{8 + 9}{6} = \frac{17}{6}$