ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2562 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{สูตร} \log \log a+\log b = \log \left(a·b\right) \\ \log a = b \to a={10}^b \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \log \left(\log x\right)+\log \left(\log x^8-16\right) = 1 \to \boxed{1} \\ \log \left[\log x\left(\log x^8-16\right)\right] = 1 \\ \log x\left(\log x^8-16\right) = {10}^1 \\ \log x\left(8\log x-16\right)-10 = 0 \end{gathered}$$

$$\begin{gathered} - \mathrm{กำหนดให้} \log x=a \\ \mathrm{จะได้} a\left(8a-16\right)-10 = 0 \\ 8a^2-16a-10 = 0 \\ 4a^2-8a-5 = 0 \\ \left(2a-5\right)\left(2a+1\right) = 0 \\ \mathrm{จะได้} 2a-5 = 0 \mathrm{หรือ} 2a+1 = 0 \\ a = \frac{5}{2} a = \frac{-1}{2} \end{gathered}$$

$$\begin{gathered} - \mathrm{แทน} a=\log x \\ \mathrm{จะได้} \log x=\frac{5}{2} \log x=\frac{-1}{2} \\ \mathrm{จาก} \boxed{1} \mathrm{นิยาม} \log \left(\log x \to x>0 \mathrm{เสมอ}\right) \\ \mathrm{จะได้} \log x>0 \mathrm{และ} \log x^8-16>0 \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} \log x=\frac{5}{2} \mathrm{เท่านั้น} \\ x = {10}^{\frac{5}{2}} \\ x = \sqrt{{10}^5} = {10}^2\sqrt{10} \\ \mathrm{ดังนั้น} x = 100\sqrt{10} \end{gathered}$$