ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2563 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากคุณสมบัติของ} Log \\ {\log }_a{x}^n = n{\log }_{a}x \\ {\log }_a\left(\frac{M}{N}\right) = {\log }_{a}M-{\log }_{a}N \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} x^{{\log }_5{x}^2} = \frac{25}{x^2} \\ \mathrm{ใส่} {\log }_5 \mathrm{ทั้งสองฝั่ง} \\ {\log }_5{x}^{{\log }_5{x}^2} = {\log }_5\left(\frac{25}{x^3}\right) \\ \left({\log }_5{x}^2\right)\left({\log }_{5}x\right) = {\log }_{5}25-{\log }_5{x}^3 \\ \left(2{\log }_{5}x\right)\left({\log }_{5}x\right) = 2-3{\log }_{5}x \end{gathered}$$

$$\begin{gathered} \mathrm{ให้} a = {\log }_{5}x \\ \left(2a^2\right) = 2-3a \\ 2a^2+3a-2 = 0 \\ \left(2a-1\right)\left(a+2\right) = 0 \\ a = \frac{1}{2} , -2 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} {\log }_{5}x = \frac{1}{2} , -2 \\ x = 5^{\frac{1}{2}} , 5^{-2} \\ x = \sqrt{5} , \frac{1}{25} \\ \mathrm{ดังนั้น} \mathrm{ผลคูณ} = \frac{\sqrt{5}}{25} \mathrm{ตอบ} 1 \end{gathered}$$