Bài 32. Hãy tính:
a) \({\log _8}12 – {\log _8}15 + {\log _8}20;\)
b) \({1 \over 2}{\log _7}36 – {\log _7}14 – 3{\log _7}\root 3 \of {21} ;\)
c) \({{{{\log }_5}36 – {{\log }_5}12} \over {{{\log }_5}9}};\)
d) \({36^{{{\log }_6}5}} + {10^{1 – \log 2}} – {8^{{{\log }_2}3}}.\)
a) \({\log _8}12 – {\log _8}15 + {\log _8}20 = {\log _8}{{12.20} \over {15}} = {\log _8}16 = {\log _{{2^3}}}{2^4} = {4 \over 3}\)
b) \({1 \over 2}{\log _7}36 – {\log _7}14 – 3{\log _7}\root 3 \of {21} = {\log _7}6 – {\log _7}14 – {\log _7}21\)
\( = {\log _7}{6 \over {14.21}} = {\log _7}{1 \over {49}} = {\log _7}{7^{ – 2}} = – 2\)
c) \({{{{\log }_5}36 – {{\log }_5}12} \over {{{\log }_5}9}} = {{{{\log }_5}{{36} \over {12}}} \over {{{\log }_5}{3^2}}} = {{{{\log }_5}3} \over {2{{\log }_5}3}} = {1 \over 2}\)
d) \({36^{{{\log }_6}5}} + {10^{1 – \log 2}} – {8^{{{\log }_2}3}} = {6^{2{{\log }_6}5}} + {10^{{{\log }_{10}}{{10} \over 2}}} – {2^{{{\log }_2}27}} = {6^{{{\log }_6}{5^2}}} + {10^{{{\log }_{10}}5}} – {2^{{{\log }_2}27}}=25 + 5 – 27 = 3\)