Question

Помогите Срочно срешенем

Answer 1

$log_{16}log_24=log_{16}(log_22^2)=log_{16}(2underbrace {log_22}_{1})=log_{16}2=\\=log_{2^4}2= frac{1}{4}log_22= frac{1}{4}\\\ frac{16}{3^{log_35}} ={a^{log_{a}b}=b}= frac{16}{5}=3,2\\\ log_{frac{1}{6}}sqrt6=log_{6^{-1}}6^{ frac{1}{2} }=- frac{1}{2}underbrace {log_66}_{1}= -frac{1}{2}$