Question

Решите 2 и 3 ..........

Answer 1

$64^{-log_{frac{1}{3}}2cdotlog_{frac{1}{4}}9+1,5}=64^{-log_{frac{1}{3}}2cdotlog_{frac{1}{4}}9}cdot64^{1,5}=\ left(64^{-log_{frac{1}{4}}9}right)^{log_{frac{1}{3}}2}cdot64^{frac{3}{2}}= left(64^{log_{4}9}right)^{log_{frac{1}{3}}2}cdot 8^3=\ left(4^{log_{4}9}right)^{3log_{frac{1}{3}}2}cdot 2^9=9^{-3log_{3}2}cdot 2^9=3^{-6log_{3}2}cdot 2^9=2^{-6}cdot 2^9=2^3=8.$

$log_364cdotlog_2frac{1}{27} =6log_32cdotleft(-3log_23right)=6cdot(-3)=-24.$