Question

с логарифмами, очень сдать нужно ЗА ПРАВИЛЬНОСТЬ

Answer 1

$1)log_{2} 5*log_{25}8=\frac{1}{log_{5}2}*log_{5}2^{3} =\frac{3log_{5}2}{log_{5}2}=3\\\\2)log_{2}log_{3} \sqrt[16]{3}=log_{2}log_{3}3^{\frac{1}{16}}=log_{2}\frac{1}{16}=log_{2} 2^{-4}=-4\\\\3)log_{\frac{1}{5}} (3\sqrt{3}+\sqrt{2} )+log_{\frac{1}{5}}(3\sqrt{3}-\sqrt{2} )=log_{\frac{1}{5}}(3\sqrt{3}+\sqrt{2})(3\sqrt{3}-\sqrt{2})=\\=log_{\frac{1}{5}}[(3\sqrt{3})^{2}-(\sqrt{2})^{2}]=log_{\frac{1}{5}}(27-2)=log_{\frac{1}{5} }25=log_{\frac{1}{5}}(\frac{1}{5})^{-2} =-2$

$4)log_{4\sqrt[4]{2}}16\sqrt{2}=log_{(2^{2}*2^{\frac{1}{4}})}2^{4}*2^{\frac{1}{2}}=log_{2^{\frac{9}{4}}} 2^{\frac{9}{2}}=log_{(2^{\frac{9}{4}})^{\frac{4}{9}}}(2^{\frac{9}{2}})^{\frac{4}{9}}=log_{2}2^{2}=2\\\\5)121^{0,5log_{11} 0,25}=11^{log_{11}0,25 }=0,25$