Đặt ${\log _2}60 = a;{\log _5}15 = b.$ Tính $P = {\log _2}12$ theo $a$ và $b$.
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A.
$P = \dfrac{{ab + 2a + 2}}{b}$
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B.
$P = \dfrac{{ab - a + 2}}{b}$
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C.
$P = \dfrac{{ab + a - 2}}{b}$
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D.
$P = \dfrac{{ab - a - 2}}{b}$
Sử dụng các công thức logarit.
\(a = {\log _2}60 = {\log _2}\left( {{2^2}.15} \right) = {\log _2}{2^2} + {\log _2}15 = 2 + {\log _2}15 \Rightarrow {\log _2}15 = a - 2\)
\( \Rightarrow {\log _2}5 = \frac{{{{\log }_{15}}5}}{{{{\log }_{15}}2}} = \frac{{{{\log }_{15}}5}}{{\frac{1}{{{{\log }_2}15}}}} = \frac{{{{\log }_2}15}}{{{{\log }_5}15}} = \frac{{a - 2}}{b}\)
\(b = {\log _5}15 = {\log _5}\left( {3.5} \right) = 1 + {\log _5}3 \Rightarrow {\log _5}3 = b - 1\)
\({\log _2}3 = {\log _2}5.{\log _5}3 = \frac{{a - 2}}{b}.\left( {b - 1} \right) = \frac{{ab - 2b - a + 2}}{b}\)
\({\log _2}12 = {\log _2}\left( {{2^2}.3} \right) = {\log _2}{2^2} + {\log _2}3 = 2 + {\log _2}3 = \frac{{ab - a + 2}}{b}\)
Đáp án : B
