ข้อสอบ PAT 1 - กุมภาพันธ์ 2563

$$\begin{gathered} \mathrm{ให้} 2^a = x \mathrm{และ} {\log }_{2}b = y \\ \mathrm{จะได้} \frac{2^a-{\log }_{2}b}{2{\log }_{2}b-4} = \frac{1}{2} \\ \frac{x-y}{2y-4} = \frac{1}{2} \\ x = 2y-2 \dots \overline{)1} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \frac{3+{\log }_{2}b}{2^a+4} = \frac{{\log }_{2}b}{2^a} \\ \frac{3+y}{x+4} = \frac{y}{x} \\ 3x = 4y \end{gathered}$$

$$\begin{gathered} \mathrm{เอา} \overline{)1} \mathrm{มาแทน} 3\left(2y-2\right) = 4y \\ y = 3 \mathrm{แทนใน} \overline{)1} \\ x = 4 \\ 2^a = 4 \end{gathered}$$

$$\begin{gathered} a = 2 \\ {\log }_{2}b = 3 \\ b = 8 \\ \mathrm{ดังนั้น} a^2+b^2 = 2^2+8^2 = 68 \mathrm{ตอบ} 5 \end{gathered}$$