ข้อสอบ PAT 1 - พฤศจิกายน 2557

$$\begin{gathered} \mathrm{จากโจทย์} {\log }_{\mathrm{a}}4+{\log }_{\mathrm{b}}4=9{\log }_{\mathrm{ab}}2 \\ {\log }_{\mathrm{a}}{2}^2+{\log }_{\mathrm{b}}{2}^2=9{\log }_{\mathrm{ab}}2 \\ 2{\log }_{\mathrm{a}}2+2{\log }_{\mathrm{b}}2=9{\log }_{\mathrm{ab}}2 \\ \frac{2\log 2}{\mathrm{loga}}+\frac{2\log 2}{\mathrm{logb}}=\frac{9\log 2}{\mathrm{logab}} \\ \mathbf{โดย} \log \mathrm{ab}=\mathrm{loga}+\mathrm{logb} ; \\ \frac{2\log 2}{\mathrm{loga}}+\frac{2\log 2}{\mathrm{logb}}=\frac{9\log 2}{\mathrm{loga}+\mathrm{logb}} \end{gathered}$$

$$\begin{gathered} \mathrm{กำหนดให้} \log \mathrm{a}=\mathrm{x} , \log \mathrm{b}=\mathrm{y} \mathrm{แทนค่าในสมการ} \\ \mathrm{จะได้} \frac{2\log 2}{\mathrm{x}}+\frac{2\log 2}{\mathrm{y}}=\frac{9\log 2}{\mathrm{x}+\mathrm{y}} \\ \cancel{\log 2}(\frac{2}{\mathrm{x}}+\frac{2}{\mathrm{y}})=\frac{9\cancel{\log 2}}{\mathrm{x}+\mathrm{y}} \\ \mathrm{ตัด} \log 2 \frac{2}{\mathrm{x}}+\frac{2}{\mathrm{y}}=\frac{9}{\mathrm{x}+\mathrm{y}} \\ \frac{2\mathrm{y}+2\mathrm{x}}{\mathrm{xy}}=\frac{9}{\mathrm{x}+\mathrm{y}} \\ \frac{2\left(\mathrm{x}+\mathrm{y}\right)}{\mathrm{xy}}=\frac{9}{\mathrm{x}+\mathrm{y}} \end{gathered}$$

$$\begin{gathered} 2{\left(\mathrm{x}+\mathrm{y}\right)}^2=9\mathrm{xy} \\ 2{\mathrm{x}}^2+4\mathrm{xy}+2{\mathrm{y}}^2=9\mathrm{xy} \\ 2{\mathrm{x}}^2-5\mathrm{xy}+2{\mathrm{y}}^2=0 \\ \left(2\mathrm{x}-\mathrm{y}\right)\left(\mathrm{x}-2\mathrm{y}\right)=0 \\ \mathrm{จะได้} 2\mathrm{x}-\mathrm{y}=0 \mathrm{หรือ} \mathrm{x}-2\mathrm{y}=0 \\ 2\mathrm{x}=\mathrm{y} \mathrm{x}=2\mathrm{y} \end{gathered}$$

$$\begin{gathered} \mathrm{แทนค่า} \mathrm{x}=\log \mathrm{a} , \mathrm{y}=\log \mathrm{b} \\ 2\log \mathrm{a}=\log \mathrm{b} \log \mathrm{a}=2\log \mathrm{b} \\ \frac{\log \mathrm{a}}{\log \mathrm{b}}=\frac{1}{2}\to \boxed{1} \frac{\log \mathrm{a}}{\log \mathrm{b}}=2\to \boxed{2} \end{gathered}$$​

$$\begin{gathered} \mathrm{หาค่ามากสุด} {\log }_{\mathrm{a}}\left({\mathrm{ab}}^5\right)+{\log }_{\mathrm{b}}\left(\frac{{\mathrm{a}}^2}{\sqrt{\mathrm{b}}}\right) \\ =\left({\log }_{\mathrm{a}}\mathrm{a}+{\log }_{\mathrm{a}}{\mathrm{b}}^5\right)+\left({\log }_{\mathrm{b}}{\mathrm{a}}^2-{\log }_{\mathrm{b}}\sqrt{\mathrm{b}}\right) \\ =\left(1+5{\log }_{\mathrm{a}}\mathrm{b}\right)+\left(2{\log }_{\mathrm{b}}\mathrm{a}-{\log }_{\mathrm{b}}{\mathrm{b}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right) \\ =(1+5\frac{\log \mathrm{b}}{\log \mathrm{a}})+(2\frac{\log \mathrm{a}}{\log \mathrm{b}}-\frac{1}{2}{\log }_{\mathrm{b}}\mathrm{b}) \\ =(1+5\frac{\log \mathrm{b}}{\log \mathrm{a}})+(2\frac{\log a}{\log b}-\frac{1}{2}) \\ \mathbf{-}\mathbf{จาก}\mathbf{ }\boxed{\mathbf{1}}\mathbf{ }\mathbf{แทนค่า}\mathbf{ }\frac{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{ }\mathbf{a}}{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{ }\mathbf{b}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \end{gathered}$$

$$\begin{gathered} {\log }_{\mathrm{a}}\left({\mathrm{ab}}^5\right)+{\log }_{\mathrm{b}}\left(\frac{{\mathrm{a}}^2}{\sqrt{\mathrm{b}}}\right)=(1+5\frac{\log \mathrm{b}}{\log \mathrm{a}})+(2\frac{\log \mathrm{a}}{\log \mathrm{b}}-\frac{1}{2}) \\ =\left[1+5\left(2\right)\right]+\left[2\right(\frac{1}{2})-\frac{1}{2}] \\ =11+0.5 \\ =11.5 \\ \mathbf{-}\mathbf{จาก}\mathbf{ }\boxed{\mathbf{2}}\mathbf{ }\mathbf{แทนค่า}\mathbf{ }\frac{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{ }\mathbf{a}}{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{ }\mathbf{b}}\mathbf{=}\mathbf{2} \end{gathered}$$

$$\begin{gathered} {\log }_{\mathrm{a}}\left({\mathrm{ab}}^5\right)+{\log }_{\mathrm{b}}\left(\frac{{\mathrm{a}}^2}{\sqrt{\mathrm{b}}}\right)=(1+5\frac{\log \mathrm{b}}{\log \mathrm{a}})+(2\frac{\log \mathrm{a}}{\log \mathrm{b}}-\frac{1}{2}) \\ =[1+5(\frac{1}{2}\left)\right]+\left[2\right(2)-\frac{1}{2}] \\ =3.5+3.5 \\ =7 \\ \mathrm{ดังนั้น} \mathrm{ค่ามากสุดของ} {\log }_{\mathrm{a}}\left({\mathrm{ab}}^5\right)+{\log }_{\mathrm{b}}\left(\frac{{\mathrm{a}}^2}{\sqrt{\mathrm{b}}}\right)=11.5 \end{gathered}$$