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1)
![log_{0.5}27=log_{ frac{1}{2} }27=log_{2^{-1}}27=-log_{2}27= \ \
=-log_{2}3^3=-3log_{2}3 \ \ \
-3log_{2}3=a \
log_{2}3=- frac{a}{3}](https://tex.z-dn.net/?f=log_%7B0.5%7D27%3Dlog_%7B+frac%7B1%7D%7B2%7D+%7D27%3Dlog_%7B2%5E%7B-1%7D%7D27%3D-log_%7B2%7D27%3D+%5C++%5C+%0A%3D-log_%7B2%7D3%5E3%3D-3log_%7B2%7D3+%5C++%5C++%5C+%0A-3log_%7B2%7D3%3Da+%5C+%0Alog_%7B2%7D3%3D-+frac%7Ba%7D%7B3%7D+ "log_{0.5}27=log_{ frac{1}{2} }27=log_{2^{-1}}27=-log_{2}27= \ \
=-log_{2}3^3=-3log_{2}3 \ \ \
-3log_{2}3=a \
log_{2}3=- frac{a}{3}")
2)
![log_{ sqrt{3} } sqrt[6]{1.5}=log_{3^{ frac{1}{2} }}(1.5^{ frac{1}{6} })= frac{1}{6}: frac{1}{2} log_{3}1.5= \ \
= frac{1}{3}log_{3}( frac{3}{2} )= frac{1}{3}(log_{3}3-log_{3}2)= \ \
= frac{1}{3}(1-log_{3}2)= frac{1}{3}(1- frac{1}{log_{2}3} )](https://tex.z-dn.net/?f=log_%7B+sqrt%7B3%7D+%7D+sqrt%5B6%5D%7B1.5%7D%3Dlog_%7B3%5E%7B+frac%7B1%7D%7B2%7D+%7D%7D%281.5%5E%7B+frac%7B1%7D%7B6%7D+%7D%29%3D+frac%7B1%7D%7B6%7D%3A+frac%7B1%7D%7B2%7D+log_%7B3%7D1.5%3D+%5C++%5C+%0A%3D+frac%7B1%7D%7B3%7Dlog_%7B3%7D%28+frac%7B3%7D%7B2%7D+%29%3D+frac%7B1%7D%7B3%7D%28log_%7B3%7D3-log_%7B3%7D2%29%3D+%5C++%5C+%0A%3D+frac%7B1%7D%7B3%7D%281-log_%7B3%7D2%29%3D+frac%7B1%7D%7B3%7D%281-+frac%7B1%7D%7Blog_%7B2%7D3%7D+%29++++++ "log_{ sqrt{3} } sqrt[6]{1.5}=log_{3^{ frac{1}{2} }}(1.5^{ frac{1}{6} })= frac{1}{6}: frac{1}{2} log_{3}1.5= \ \
= frac{1}{3}log_{3}( frac{3}{2} )= frac{1}{3}(log_{3}3-log_{3}2)= \ \
= frac{1}{3}(1-log_{3}2)= frac{1}{3}(1- frac{1}{log_{2}3} )")
3)
Ответ: 3)
2)
3)
Ответ: 3)
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