Câu hỏi 6:

Tính:$$\log _{\frac{1}{2}} 2+2 \log _{\frac{1}{2}} \frac{1}{3}+\log _{\frac{1}{2}} \frac{3}{8}$$

Lời giải:

Ta có:

$$\begin{aligned}& \log _{\frac{1}{2}} 2+2 \log _{\frac{1}{2}} \frac{1}{3}+\log _{\frac{1}{2}} \frac{3}{8} \\& =\log _{\frac{1}{2}} 2+\log _{\frac{1}{2}} \frac{1}{3}+\log _{\frac{1}{2}} \frac{1}{3}+\log _{\frac{1}{2}} \frac{3}{8} \\& =\log _{\frac{1}{2}}\left(2 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{3}{8}\right)=\log _{\frac{1}{2}} \frac{1}{12}\end{aligned}$$