Câu hỏi:
2 năm trước
Đặt \(a = {\log _{27}}5,\,\,b = {\log _8}7,\,\,c = {\log _2}3\). Khi đó \({\log _{12}}35\) bằng:
Trả lời bởi giáo viên
Đáp án đúng: c
Ta có:
\(\begin{array}{l}\,\,\,{\log _{12}}35 = {\log _{12}}\left( {5.7} \right)\\ = {\log _{12}}5 + {\log _{12}}7\\ = \dfrac{1}{{{{\log }_5}\left( {{2^2}.3} \right)}} + \dfrac{1}{{{{\log }_7}\left( {{2^2}.3} \right)}}\\ = \dfrac{1}{{2{{\log }_5}2 + {{\log }_5}3}} + \dfrac{1}{{2{{\log }_7}2 + {{\log }_7}3}}\end{array}\)
Theo bài ra ta có:
\(\begin{array}{l}a = {\log _{27}}5 = \dfrac{1}{{{{\log }_5}{3^3}}} = \dfrac{1}{{3{{\log }_5}3}}\\ \Rightarrow {\log _5}3 = \dfrac{1}{{3a}}\\b = {\log _8}7 = \dfrac{1}{{{{\log }_7}{2^3}}} = \dfrac{1}{{3{{\log }_7}2}}\\ \Rightarrow {\log _7}2 = \dfrac{1}{{3b}}\\c = {\log _2}3 \Rightarrow {\log _3}2 = \dfrac{1}{c}\\ \Rightarrow {\log _5}2 = {\log _5}3.{\log _3}2 = \dfrac{1}{{3a}}.\dfrac{1}{c} = \dfrac{1}{{3ac}}\\\,\,\,\,\,\,{\log _7}3 = {\log _7}2.{\log _2}3 = \dfrac{1}{{3b}}.c = \dfrac{c}{{3b}}\end{array}\)
Khi đó ta có:
\(\begin{array}{l}\,\,\,{\log _{12}}35 = {\log _{12}}\left( {5.7} \right)\end{array}\)
$= \dfrac{1}{{\dfrac{2}{{3ac}} + \dfrac{1}{{3a}}}} + \dfrac{1}{{\dfrac{2}{{3b}} + \dfrac{c}{{3b}}}}$
$= \dfrac{{3ac}}{{2 + c}} + \dfrac{{3b}}{{2 + c}} = \dfrac{{3ac + 3b}}{{c + 2}}$
Hướng dẫn giải:
Sử dụng các công thức:
\(\begin{array}{l}{\log _a}\left( {xy} \right) = {\log _a}x + {\log _a}y\,\,\left( {0 < a \ne 1,\,\,x,y > 0} \right)\\{\log _a}b.{\log _b}c = {\log _a}c\,\,\left( {0 < a,b \ne 1,\,\,c > 0} \right)\\{\log _a}b = \dfrac{1}{{{{\log }_b}a}}\,\,\left( {0 < a,b \ne 1} \right)\end{array}\)
