Bài tập 9 trang 13 SBT Toán 11 Tập 2 Chân trời sáng tạo - CTST

Ta có:

• \(a={{\log }_{2}}3=\frac{1}{{{\log }_{3}}2}\Rightarrow {{\log }_{3}}2=\frac{1}{a};\)
• \(\begin{align} & b={{\log }_{3}}15={{\log }_{3}}(3.5)={{\log }_{3}}3+{{\log }_{3}}5=1+{{\log }_{3}}5 \\ & \Rightarrow {{\log }_{3}}5=b-1 \\ \end{align}\)

Suy ra:

\(\begin{align} & {{\log }_{30}}18=\frac{{{\log }_{3}}18}{{{\log }_{3}}30}=\frac{{{\log }_{30}}({{2.3}^{2}})}{{{\log }_{3}}(2.3.5)}=\frac{{{\log }_{3}}2+{{\log }_{3}}{{3}^{2}}}{{{\log }_{3}}2+{{\log }_{3}}3+{{\log }_{3}}5} \\ & =\frac{{{\log }_{3}}2+2}{{{\log }_{3}}2+1+{{\log }_{3}}5}=\frac{\frac{1}{a}+2}{\frac{1}{a}+a+b-1} \\ & =\frac{2a+1}{ab+1} \\ \end{align}\)

Vậy \({{\log }_{30}}18=\frac{2a+1}{ab+1}\).

-- Mod Toán 11 HỌC247