ข้อสอบ PAT 1 - กุมภาพันธ์ 2562

$$\begin{gathered} \mathrm{จากโจทย์} {\left({\log }_{3}9x\right)}^2-3{\log }_{\sqrt{3}}x-7 = 0 \to \boxed{1} \\ {\left({\log }_{3}9+{\log }_{3}x\right)}^2-3{\log }_{\sqrt{3}}x-7 = 0 \\ {\left({\log }_3{3}^2+{\log }_{3}x\right)}^2-3\left(\frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right){\log }_{3}x-7 = 0 \\ {\left(2{\log }_{3}3+{\log }_{3}x\right)}^2-3\left(2\right){\log }_{3}x-7 = 0 \\ {\left(2+{\log }_{3}x\right)}^2-6{\log }_{3}x-7 = 0 \to \boxed{2} \end{gathered}$$

$$\begin{gathered} \mathrm{กำหนดให้} A = {\log }_{3}x \\ \mathrm{จาก} \boxed{2} \mathrm{จะได้} {\left(2+A\right)}^2-6A-7 = 0 \\ 4+4A+A^2-6A-7 = 0 \\ A^2-2A-3 = 0 \\ \left(A-3\right)\left(A+1\right) = 0 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} A-3 = 0 \mathrm{หรือ} A+1 = 0 \\ A = 3 A = -1 \\ {\log }_{3}x = 3 {\log }_{3}x = -1 \\ x = 3^3 x = 3^{-1} \\ x = 27 x = \frac{1}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{เงื่อนไข} \mathrm{นิยาม} {\log }_{a}m \to m>0 \\ \mathrm{จาก} \boxed{1} \mathrm{จะได้} 9x > 0 \mathrm{และ} x>0 \\ x > 0 \\ \mathrm{แสดงว่า} x = 27,\frac{1}{3} \\ \mathrm{ดังนั้น} \mathrm{ผลคูณของสมาชิกทั้งหมด} = 27\times \frac{1}{3} \\ = 9 \end{gathered}$$