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Решите задачу:

$log_{\frac{1}{\sqrt{2}}}(2\sqrt{2})=log_{\frac{1}{2^{\frac{1}{2}}}}(2^1*2^{\frac{1}{2}})=log_{2^{-\frac{1}{2}}}(2^{\frac{3}{2}})=\\\\ =\frac{3}{2}*log_{2^{-\frac{1}{2}}}(2)=\frac{3}{2}*\frac{1}{-\frac{1}{2}}*log_{2}}(2)=\\\\ =\frac{3}{2}*(-2)*1=-3\\\\ ----------------\\ log_{4}\frac{1}{2\sqrt{2}}=log_{2^2}(2^{-\frac{3}{2}})=-\frac{3}{2}*\frac{1}{2}*log_2(2)=-\frac{3}{4}\\\\ -----------------\\ log_{\sqrt{27}}(9)=log_{3^{\frac{3}{2}}}(3^2)=\frac{2}{3}*2*log_3(3)=\frac{4}{3}\\\\ ----------------\\$

$log_{16}(\sqrt{2})=log_{2^4}(2^{\frac{1}{2}})=\frac{1}{4}*\frac{1}{2}*log_{2}(2)=\frac{1}{8}\\\\ -------------------\\ 27^{log_3(2)}=(27)^{log_3(2)}=(3^3)^{log_3(2)}=3^{3*log_3(2)}=3^{log_3(2^3)}=\\ =3^{log_3(8)}=8\\\\ --------------------\\ 9^{-log_{3}(4)}=3^{-2log_3(4)}=3^{log_3(4^{-2})}=4^{-2}=16^{-1}=\frac{1}{16}\\\\ -------------------\\ \frac{log_3(4)+log_3(5)}{log_9(2)}=\frac{log_3(4*5)}{log_{3^2}(2)}=\frac{log_3(20)}{\frac{1}{2}*log_{3}(2)}=2*\frac{log_3(20)}{log_3(2)}=\\ =2*log_2(20)\\\\$

$\frac{log_2(9)}{log_2(5)*log_5(3)}=\frac{log_2(9)}{log_2(5)}*\frac{1}{log_5(3)}}=log_5(9)*\frac{1}{log_5(3)}}=\\ =\frac{log_5(9)}{log_5(3)}}=log_3(9)=2$