Bài tập nâng cao · Bài 4
Bài tập nâng cao Lôgarit · Bài 4
Cho $\log_{9}5=a;\,\,\log_{4}7=b;\,\,\log_{2}3=c$. Biết $\log_{24}175=\frac{mb+nac}{pc+q}$. Tính $A=m+2n+3p+4q$.
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Lời giải
Ta có $\log_{24}175=\log_{24}7.5^{2}=\log_{24}7+2\log_{24}5=\frac{1}{\log_{7}24}+\frac{2}{\log_{5}24}=$
$\frac{1}{\log_{7}3+\log_{7}2^{3}}+\frac{2}{\log_{5}3+\log_{5}2^{3}}=\frac{1}{\frac{1}{\log_{3}7}+\frac{3}{\log_{2}7}}+\frac{2}{\frac{1}{\log_{3}5}+\frac{3}{\log_{2}5}}=$
$\frac{1}{\frac{1}{\log_{2}7.\log_{3}2}+\frac{3}{\log_{2}7}}+\frac{2}{\frac{1}{\log_{3}5}+\frac{3}{\log_{2}3.\log_{3}5}}=\frac{1}{\frac{1}{2b.\frac{1}{c}}+\frac{3}{2b}}+\frac{2}{\frac{1}{2a}+\frac{3}{c.2a}}=$
$\frac{1}{\frac{c}{2b}+\frac{3}{2b}}+\frac{2}{\frac{c}{2ac}+\frac{3}{2ac}}=\frac{2b}{c+3}+\frac{4ac}{c+3}=\frac{2b+4ac}{c+3}$.
$A=m+2n+3p+4q=2+8+3+12=25$
$\frac{1}{\log_{7}3+\log_{7}2^{3}}+\frac{2}{\log_{5}3+\log_{5}2^{3}}=\frac{1}{\frac{1}{\log_{3}7}+\frac{3}{\log_{2}7}}+\frac{2}{\frac{1}{\log_{3}5}+\frac{3}{\log_{2}5}}=$
$\frac{1}{\frac{1}{\log_{2}7.\log_{3}2}+\frac{3}{\log_{2}7}}+\frac{2}{\frac{1}{\log_{3}5}+\frac{3}{\log_{2}3.\log_{3}5}}=\frac{1}{\frac{1}{2b.\frac{1}{c}}+\frac{3}{2b}}+\frac{2}{\frac{1}{2a}+\frac{3}{c.2a}}=$
$\frac{1}{\frac{c}{2b}+\frac{3}{2b}}+\frac{2}{\frac{c}{2ac}+\frac{3}{2ac}}=\frac{2b}{c+3}+\frac{4ac}{c+3}=\frac{2b+4ac}{c+3}$.
$A=m+2n+3p+4q=2+8+3+12=25$