ข้อสอบ PAT 1 - กุมภาพันธ์ 2563

$$\begin{gathered} {\log }_2\left(2^{\sqrt{x}}+{\left(2x\right)}^{\log x}-4^{\log 8}\right) = {\left(\sqrt{2}\right)}^{{\log }_{2}x} \\ {\log }_2\left(2^{\sqrt{x}}+{\left(2x\right)}^{\log x}-4^{\log 8}\right) = \sqrt{2^{{\log }_{2}x}} \\ {\log }_2\left(2^{\sqrt{x}}+{\left(2x\right)}^{\log x}-4^{\log 8}\right) = \sqrt{x} \\ 2^{\sqrt{x}}+{\left(2x\right)}^{\log x}-4^{\log 8} = 2^{\sqrt{x}} \end{gathered}$$

$$\begin{gathered} {\left(2x\right)}^{\log x} = 4^{\log 8} \\ \log {\left(2x\right)}^{\log x} = \log 4^{\log 8} \\ \left(\log x\right)\left(\log 2+\log x\right) = \left(\log 8\right)\left(\log 4\right) \\ \mathrm{เปลี่ยนตัวแปล} \mathrm{ให้} a=\log x \\ \left(a\right)\left(\log 2+a\right) = \left(\log 8\right)\left(\log 4\right) \end{gathered}$$

$$\begin{gathered} a^2+\left(\log 2\right)a - \left(\log 8\right)\left(\log 4\right) = 0 \\ \left(a+\log 8\right)\left(a-\log 4\right) = 0 \\ a = -\log 8 , \log 4 \\ \log x = \log 8^{-1} , \log 4 \\ x = \frac{1}{8} , 4 \\ \mathrm{ดังนั้น} \mathrm{ผลคูณ} = \left(\frac{1}{8}\right)\left(4\right) = \frac{1}{2} = 0.5 \end{gathered}$$