Bài 3. Logarit – SBT Toán lớp 12
Bài 2.12 trang 108 SBT Giải tích 12
Tính:
a) \({(\frac{1}{9})^{\frac{1}{2}{{\log }_3}4}}\)
b) \({10^{3 – \log 5}}\)
c) \(2{\log _{27}}\log 1000\)
d) \(3{\log _2}{\log _4}16 + {\log _{\frac{1}{2}}}2\)
Hướng dẫn giải
a) \(\frac{1}{4}\)
b) \(\frac{{{{10}^3}}}{{{{10}^{\log 5}}}} = \frac{{{{10}^3}}}{5} = 200\)
c) \(\frac{2}{3}\)
d) 2
Bài 2.13 trang 108
Tính:
a) \(\frac{1}{2}{\log _7}36 – {\log _7}14 – 3{\log _7}\sqrt[3]{{21}}\)
b) \(\frac{{{{\log }_2}24 – \frac{1}{2}{{\log }_2}72}}{{{{\log }_3}18 – \frac{1}{3}{{\log }_3}72}}\)
c) \(\frac{{{{\log }_2}4 + {{\log }_2}10}}{{{{\log }_2}20 + 3{{\log }_2}2}}\)
Bài giải:
a) \({\log _7}\sqrt {36} – {\log _7}14 – {\log _7}21 = {\log _7}\frac{1}{{49}} = – 2\)
b) \(\frac{{{{\log }_2}24 – {{\log }_2}\sqrt {72} }}{{{{\log }_3}18 – {{\log }_3}\sqrt[3]{{72}}}} = \frac{{{{\log }_2}{2^{\frac{3}{2}}}}}{{{{\log }_3}{3^{\frac{4}{3}}}}} = \frac{9}{8}\)
c) \(\frac{{{{\log }_2}24 – {{\log }_2}\sqrt {72} }}{{{{\log }_3}18 – {{\log }_3}\sqrt[3]{{72}}}} = \frac{{{{\log }_2}{2^{\frac{3}{2}}}}}{{{{\log }_3}{3^{\frac{4}{3}}}}} = \frac{9}{8}\).
Bài 2.14 trang 108
Tìm x, biết:
a) \({\log _5}x = 2{\log _5}a – 3{\log _5}b\)
b) \({\log _{\frac{1}{2}}}x = \frac{2}{3}{\log _{\frac{1}{2}}}a – \frac{1}{5}{\log _{\frac{1}{2}}}b\)
Hướng dẫn giải
a) \(x = \frac{{{a^2}}}{{{b^3}}}\)
b) \(x = \frac{{{a^{\frac{2}{3}}}}}{{{b^{\frac{1}{5}}}}}\).
Bài 2.15 trang 108
a) Cho \(a = {\log _3}15,b = {\log _3}10\) . Hãy tính \({\log _{\sqrt 3 }}50\) theo a và b.
b) Cho \(a = {\log _2}3,b = {\log _3}5,c = {\log _7}2\) . Hãy tính \({\log _{140}}63\) theo a, b, c.
Bài giải
a) Ta có:
\(a = {\log _3}15 = {\log _3}(3.5) = {\log _3}3 + {\log _3}5 = 1 + {\log _3}5\)
Suy ra \({\log _3}5 = a – 1\)
\(b = {\log _3}10 = {\log _3}(2.5) = {\log _3}2 + {\log _3}5\)
Suy ra \({\log _3}2 = b – {\log _3}5 = b – (a – 1) = b – a + 1\)
Do đó:
\({\log _{\sqrt 3 }}50 = {\log _{{3^{\frac{1}{2}}}}}({2.5^2}) = 2{\log _3}2 + 4{\log _3}5 = 2(b – a + 1) + 4(a – 1) = 2a + 2b – 2\)
b) Ta có:
\(\begin{array}{l}
{\log _{140}}63 = {\log _{140}}({3^2}.7) = 2{\log _{140}}3 + {\log _{140}}7\\
= \frac{2}{{{{\log }_3}140}} + \frac{1}{{{{\log }_7}140}} = \frac{2}{{{{\log }_3}({2^2}.5.7)}} + \frac{1}{{{{\log }_7}({2^2}.5.7)}}\\
= \frac{2}{{2{{\log }_3}2 + {{\log }_3}5 + {{\log }_3}7}} + \frac{1}{{2{{\log }_7}2 + {{\log }_7}5 + 1}}
\end{array}\)
Từ đề bài suy ra:
\(\begin{array}{l}
{\log _3}2 = \frac{1}{{{{\log }_2}3}} = \frac{1}{a}\\
{\log _{\frac{1}{2}}}\pi {\log _7}5 = {\log _7}2.{\log _2}3.{\log _3}5 = cab\\
{\log _3}7 = \frac{1}{{{{\log }_7}3}} = \frac{1}{{{{\log }_7}2.{{\log }_2}3}} = \frac{1}{{ca}}
\end{array}\)
Vậy \({\log _{140}}63 = \frac{2}{{\frac{2}{a} + b + \frac{1}{{ca}}}} + \frac{1}{{2c + cab + 1}} = \frac{{2ac + 1}}{{abc + 2c + 1}}\).
Bài 2.16 trang 108
Hãy so sánh mỗi cặp số sau:
a) \({\log _3}\frac{6}{5}\) và \({\log _3}\frac{5}{6}\)
b) \({\log _{\frac{1}{3}}}9\) và \({\log _{\frac{1}{3}}}17\)
c) \({\log _{\frac{1}{2}}}e\) và \({\log _{\frac{1}{2}}}\pi \)
d) \(6\pi {\log _2}\frac{{\sqrt 5 }}{2}\) và \({\log _2}\frac{{\sqrt 3 }}{2}\)
Bài giải:
a) \({\log _3}\frac{6}{5}\) > \({\log _3}\frac{5}{6}\)
b) \({\log _{\frac{1}{3}}}9\) < \({\log _{\frac{1}{3}}}17\)
c) \({\log _{\frac{1}{2}}}e\) > \({\log _{\frac{1}{2}}}\pi \)
d) \(6\pi {\log _2}\frac{{\sqrt 5 }}{2}\) > \({\log _2}\frac{{\sqrt 3 }}{2}\).
Bài 2.17
Chứng minh rằng:
a) \({\log _{{a_1}}}{a_2}.{\log _{{a_2}}}{a_3}{\log _{{a_3}}}{a_4}…..{\log _{{a_{n – 1}}}}{a_n} = {\log _{{a_1}}}{a_n}\)
b) \(\frac{1}{{{{\log }_a}b}} + \frac{1}{{{{\log }_{{a^2}}}b}} + \frac{1}{{{{\log }_{{a^3}}}b}} + … + \frac{1}{{{{\log }_{{a^n}}}b}} = \frac{{n(n + 1)}}{{2{{\log }_a}b}}\)
Bài làm
a) Sử dụng tính chất: \({\log _a}b.{\log _b}c = {\log _a}c\)
b) Sử dụng tính chất: \({\log _{{a^k}}}b = \frac{1}{k}{\log _a}b\)
và \(1 + 2 + … + n = \frac{{n(n + 1)}}{2}\)
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