Прологарифмируйте выражение (216*корень пятой степени из числа a^2)/b^3 по основанию корень из 6.
Ответ
20
Применяем формулы
![log_{ \sqrt{6} }( \frac{216 \sqrt[5]{a^2}}{b^3})=log_{ \sqrt{6} }( 216 \sqrt[5]{a^2})-log_{ \sqrt{6} }(b^3)= \\ \\ = log_{ \sqrt{6} } 216+log_{ \sqrt{6} }a^{ \frac{2}{5} }-log_{ \sqrt{6} }(b^3)= \\ \\ = \frac{1}{ \frac{1}{2} } log_6 216+ \frac{1}{ \frac{1}{2} } log_6a^{ \frac{2}{5} }- \frac{1}{ \frac{1}{2} } log_6(b^3)= \\ \\ =6+ \frac{4}{5}log_6a-6log_6b](https://tex.z-dn.net/?f=log_%7B+%5Csqrt%7B6%7D+%7D%28+%5Cfrac%7B216+%5Csqrt%5B5%5D%7Ba%5E2%7D%7D%7Bb%5E3%7D%29%3Dlog_%7B+%5Csqrt%7B6%7D+%7D%28+216+%5Csqrt%5B5%5D%7Ba%5E2%7D%29-log_%7B+%5Csqrt%7B6%7D+%7D%28b%5E3%29%3D+%5C%5C++%5C%5C+%3D+log_%7B+%5Csqrt%7B6%7D+%7D+216%2Blog_%7B+%5Csqrt%7B6%7D+%7Da%5E%7B+%5Cfrac%7B2%7D%7B5%7D+%7D-log_%7B+%5Csqrt%7B6%7D+%7D%28b%5E3%29%3D+%5C%5C++%5C%5C+%3D+%5Cfrac%7B1%7D%7B+%5Cfrac%7B1%7D%7B2%7D+%7D++log_6+216%2B+%5Cfrac%7B1%7D%7B+%5Cfrac%7B1%7D%7B2%7D+%7D+log_6a%5E%7B+%5Cfrac%7B2%7D%7B5%7D+%7D-+%5Cfrac%7B1%7D%7B+%5Cfrac%7B1%7D%7B2%7D+%7D+log_6%28b%5E3%29%3D+%5C%5C++%5C%5C+%3D6%2B+%5Cfrac%7B4%7D%7B5%7Dlog_6a-6log_6b+%0A "log_{ \sqrt{6} }( \frac{216 \sqrt[5]{a^2}}{b^3})=log_{ \sqrt{6} }( 216 \sqrt[5]{a^2})-log_{ \sqrt{6} }(b^3)= \\ \\ = log_{ \sqrt{6} } 216+log_{ \sqrt{6} }a^{ \frac{2}{5} }-log_{ \sqrt{6} }(b^3)= \\ \\ = \frac{1}{ \frac{1}{2} } log_6 216+ \frac{1}{ \frac{1}{2} } log_6a^{ \frac{2}{5} }- \frac{1}{ \frac{1}{2} } log_6(b^3)= \\ \\ =6+ \frac{4}{5}log_6a-6log_6b")
при a>0; b>0
при a>0; b>0