Bài 32 trang 92 SGK Đại số và Giải tích 12 Nâng cao

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) \({36^{{{\log }_6}5}} + {10^{1 - \log 2}} - {8^{{{\log }_2}3}}.\)

Giải

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) \({36^{{{\log }_6}5}} + {10^{1 - \log 2}} - {8^{{{\log }_2}3}} = {6^{2{{\log }_6}5}} + {10^{{{\log }_{10}}{{10} \over 2}}} - {2^{{{\log }_2}27}} = {6^{{{\log }_6}{5^2}}} + {10^{{{\log }_{10}}5}} - {2^{{{\log }_2}27}}=25 + 5 - 27 = 3\)