ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2561 (วิชาสามัญเก่า)

$$\begin{gathered} \mathbf{นิยาม}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\log }_{n}m=\frac{1}{{\log }_{m}n} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\log }_{n^b}{m}^a=\frac{a}{b}{\log }_{n}m \\ \mathrm{จากโจทย์}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{{\log }_{2}100}+\frac{1}{{\log }_5{\displaystyle 100}} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \frac{1}{{\log }_{2}100}+\frac{1}{{\log }_5{\displaystyle 100}}={\log }_{100}2+{\log }_{100}5 \\ ={\log }_{100}\left(2\times 5\right) \\ ={\log }_{100}10 \\ ={\log }_{{10}^2}{10}^1 \\ =\frac{1}{2}{\log }_{10}10 ; {\log }_{10}10=1 \\ =\frac{1}{2} \end{gathered}$$