ข้อสอบคณิต 9 สามัญ - ปี 2557

$$\begin{gathered} \mathrm{จากโจทย์} {\log }_{2}x+6{\log }_{x}2-5 = 0 \\ {\log }_{2}x+6\left(\frac{1}{{\log }_{2}x}\right)-5 = 0 \\ \mathrm{กำหนดให้} {\log }_{2}x = A \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} A+6\left(\frac{1}{A}\right)-5 = 0 \hspace{0.17em}\hspace{0.17em} ; \mathrm{นำ} A \mathrm{คูณตลอด} \\ A^2+6-5A = 0 \\ A^2-5A+6 = 0 \\ \left(A-2\right)\left(A-3\right) = 0 \\ A = 2, 3 \\ A = 2 \mathrm{หรือ} A = 3 \\ {\log }_{2}x = 2 {\log }_{2}x = 3 \\ x = 2^2 x = 2^3 \\ x = 4 x = 8 \\ \mathrm{ดังนั้น}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ผลบวกของคำตอบคือ} 4+8=12 \end{gathered}$$