02Jan

Решите 2 и 3 ..........

25071996 Алгебра 1 0

Решите 2 и 3 ..........

Posted by 25071996 | Posted at Jan 02, 2014 | Categories: Алгебра

Answers

byka
byka

[latex]64^{-log_{frac{1}{3}}2cdotlog_{frac{1}{4}}9+1,5}=64^{-log_{frac{1}{3}}2cdotlog_{frac{1}{4}}9}cdot64^{1,5}=\ left(64^{-log_{frac{1}{4}}9} ight)^{log_{frac{1}{3}}2}cdot64^{frac{3}{2}}= left(64^{log_{4}9} ight)^{log_{frac{1}{3}}2}cdot 8^3=\ left(4^{log_{4}9} ight)^{3log_{frac{1}{3}}2}cdot 2^9=9^{-3log_{3}2}cdot 2^9=3^{-6log_{3}2}cdot 2^9=2^{-6}cdot 2^9=2^3=8. [/latex] [latex]log_364cdotlog_2frac{1}{27} =6log_32cdotleft(-3log_23 ight)=6cdot(-3)=-24.[/latex]

Jan 02, 2014 23:23

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