Bài 32 trang 92 SGK Đại số và Giải tích 12 Nâng cao, Hãy tính:...

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} oot 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} oot 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)