Вычислите: 1. 2. 3.

Решите задачу:

$\frac{\log_{8}20 }{\log_{8}5 } + \log_{5}0,05 = \log_{5}20+log_{5}20^{-1} = \log_{5}20-\log_{5}20 = 0;$

$\log_{16}\log _{6}36 = \log_{2^4}\log _{6}6^2 = \frac{1}{4}log_{2}(2\log _{6}6) = \frac{1}{4}log_{2}2 = \frac{1}{4};$

$( 8^{\frac{1}{3}+\log _{2}3 } ): \log_{2}\log _{3}81=((2^3)^{\frac{1}{3}}\cdot(2^3)^{\log _{2}3 } ): \log_{2}4=2\cdot3^3:2=27.$