Đáp án:
$\begin{array}{l}
a)2 < 3 \Rightarrow {\log _2}3 > 0\\
3 > \dfrac{1}{2} \Rightarrow {\log _3}\dfrac{1}{2} < 0\\
\Rightarrow {\log _2}3 > {\log _3}\dfrac{1}{2}\\
b){3^{{{\log }_{0,1}}5}} = {5^{{{\log }_{0,1}}3}}\\
{5^{{{\log }_{0,1}}{2^{ - 2}}}} = {5^{{{\log }_{0,1}}\dfrac{1}{4}}}\\
Do:{\log _{0,1}}3 = {\log _{{{10}^{ - 1}}}}3 = - {\log _{10}}3\\
{\log _{0,1}}\dfrac{1}{4} = {\log _{{{10}^{ - 1}}}}\dfrac{1}{4} = - {\log _{10}}\dfrac{1}{4}\\
{\log _{10}}3 > {\log _{10}}\dfrac{1}{4}\\
\Rightarrow - {\log _{10}}3 < - {\log _{10}}\dfrac{1}{4}\\
\Rightarrow {5^{{{\log }_{0,1}}3}} < {5^{{{\log }_{0,1}}{2^{ - 2}}}}\\
c){\log _2}7 = \dfrac{1}{{{{\log }_7}2}}\\
{\log _3}7 = \dfrac{1}{{{{\log }_7}3}}\\
Do:{\log _7}2 < {\log _7}3\\
\Rightarrow {\log _2}7 > {\log _3}7\\
d){\log _{\dfrac{2}{3}}}5 = \dfrac{1}{{{{\log }_5}\dfrac{2}{3}}}\\
{\log _{\dfrac{3}{4}}}5 = \dfrac{1}{{{{\log }_5}\dfrac{3}{4}}}\\
{\log _5}\dfrac{2}{3} > {\log _5}\dfrac{3}{4}\\
\Rightarrow lo{g_{\dfrac{2}{3}}}5 < {\log _{\dfrac{3}{4}}}5
\end{array}$
