Nội dung bài giảng
Bài 5.
a) Cho \(a = lo{g_{30}}3,b = lo{g_{30}}5\). Hãy tính \(lo{g_{30}}1350\) theo \(a, b\).
b) Cho \(c =lo{g_{15}}3\). Hãy tính \(lo{g_{25}}15\) theo \(c\).
Giải:
a) Ta có \(1350 = 30.3^2 .5\) suy ra
\(lo{g_{30}}1350 =lo{g_{30}}(30.{3^2}.5) =1 + 2lo{g_{30}}3 + lo{g_{30}}5\)\( = 1 + 2a+b\).
b) \(lo{g_{25}}15\) = \(\frac{1}{log_{15}25}\) = \(\frac{1}{2log_{15}5}\) = \(\frac{1}{2log_{15}\left ( 15: 3 \right )}\) = \(\frac{1}{2\left (1-log_{15}3 \right )}\) = \(\frac{1}{2\left (1-c \right )}\).