Bài 2.15: 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.
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:
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\(\begin{array}{l}
{\log _{140}}63 = {\log _{140}}({3^2}.7) = 2{\log _{140}}3 + {\log _{140}}7\\
= \frac{2}{{{{\log }_3}140}} + \frac{1}{{{{\log }_7}140}} = \frac{2}{{{{\log }_3}({2^2}.5.7)}} + \frac{1}{{{{\log }_7}({2^2}.5.7)}}\\
= \frac{2}{{2{{\log }_3}2 + {{\log }_3}5 + {{\log }_3}7}} + \frac{1}{{2{{\log }_7}2 + {{\log }_7}5 + 1}}
\end{array}\)
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: Hãy so sánh mỗi cặp số sau:
a) \({\log _3}\frac{6}{5}\) và \({\log _3}\frac{5}{6}\)
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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 \({\log _2}\frac{{\sqrt 3 }}{2}\)
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) \(\frac{1}{{{{\log }_a}b}} + \frac{1}{{{{\log }_{{a^2}}}b}} + \frac{1}{{{{\log }_{{a^3}}}b}} + … + \frac{1}{{{{\log }_{{a^n}}}b}} = \frac{{n(n + 1)}}{{2{{\log }_a}b}}\)
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 \(1 + 2 + … + n = \frac{{n(n + 1)}}{2}\)
