Toán Lớp 12: Cho $log_{a}bc=2$, $log_{b}ca=3$. Tính $S=log_{c}ab$
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Toán Lớp 12: Cho $log_{a}bc=2$, $log_{b}ca=3$. Tính $S=log_{c}ab$
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$log_{c}(ab)=\frac{7}{5}$
$log_{b}ca=3⇒log_{b}b+log_{b}ca=4⇔log_{b}(abc)=4$
$log_{abc}(abc)=log_{abc}a+log_{abc}b+log_{abc}c=1$
$⇔\frac{1}{log_{a}(abc)}+\frac{1}{log_{b}(abc)}+\frac{1}{log_{c}(abc)}=1$