Question

Вариант 3 с 3 по 5 с решением

Answer 1

$3)\; log_{0,5}\sqrt[5]{64}=log_{2^{-1}}(2^{\frac{6}{5}})=-\frac{6}{5}log_22=-\frac{6}{5}\\\\4)\; \; log_{81}(log_{\sqrt[9]{2}}8)=log_{3^4}(\frac{1}{9}log_22^3)=\frac{1}{4}log_3(\frac{3}{9}log_22)=\frac{1}{4}log_3\frac{1}{3}=\\\\=\frac{1}{4}log_3(3^{-1})=-\frac{1}{4}$

$5)\; log_{a}\sqrt{b}=\frac{1}{2}log_{a}b=0,6\; \; \to \; \; log_{a}b=0,6\cdot 2=1,2\\\\log_{b}(a^6b^{99})=log_{b}a^6+log_{b}b^{99}=6log_{b}a+99log_{b}b=\frac{6}{log_{a}b}+99=\\\\=\frac{6}{1,2}+99=5+99=104$