ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} 5^{\left(\mathrm{x}-2^{\mathrm{A}}\right)}{2}^{{\mathrm{y}}^{\mathrm{A} }}={\left(16\right)}^{64} \\ 5^{\left(\mathrm{x}-2^{\mathrm{A}}\right)}{2}^{{\mathrm{y}}^{\mathrm{A} }}={\left(2^4\right)}^{64}{5}^0 \\ \mathrm{เทียบ} \mathrm{สปส}. \mathrm{ของเลขฐาน} 2 \mathrm{และฐาน} 3 \\ \mathrm{จะได้} \mathrm{x}-2^{\mathrm{A}}=0 \mathrm{และ} {\mathrm{y}}^{\mathrm{A}}=\left(4\right)\left(64\right) \\ \mathrm{x}=2^{\mathrm{A}}\to \boxed{1} {\mathrm{y}}^{\mathrm{A}}=256\to \boxed{2} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \mathrm{A}=\frac{\log \mathrm{y}}{\log \mathrm{x}} \\ \mathrm{จะได้} \mathrm{A}={\log }_{\mathrm{x}} \mathrm{y} \\ {\mathrm{x}}^{\mathrm{A}}=\mathrm{y} \\ \mathrm{y}={\mathrm{x}}^{\mathrm{A}}\to \boxed{3} \end{gathered}$$

$$\begin{gathered} \mathrm{แทน} \boxed{3} \mathrm{ใน} \boxed{2} \\ \mathrm{จะได้} {\left({\mathrm{x}}^{\mathrm{A}}\right)}^{\mathrm{A}}=256 \\ {\mathrm{x}}^{{\mathrm{A}}^2}=256 ; \mathrm{จาก} \boxed{1} \mathrm{x}=2^{\mathrm{A}} \\ {\left(2^{\mathrm{A}}\right)}^{{\mathrm{A}}^2}=256 \\ 2^{{\mathrm{A}}^3}=2^8 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} {\mathrm{A}}^3=8 \\ \mathrm{A}=2 \\ \mathrm{จาก} \boxed{1} \mathrm{จะได้} \mathrm{x}=2^2=4 \\ \mathrm{จาก} \boxed{3} \mathrm{จะได้} \mathrm{y}=4^2=16 \\ \mathrm{ดังนั้น} \mathrm{ค่าของ} \mathrm{x}+\mathrm{y}=4+16 \\ =20 \end{gathered}$$