ข้อสอบคณิต O-NET - 2549

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{จะเห็นว่า} \\ 1000, 600, 300 \mathrm{หารด้วย} 100 \mathrm{ลงตัว} \\ \mathrm{จะได้} 2^{1000} = {\left(2^{10}\right)}^{100} \\ 3^{600} = {\left(3^6\right)}^{100} \\ {10}^{300} = {\left({10}^3\right)}^{100} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} 2^{10} = 2^3·2^3·2^4 = 8·8·16 = 1024 \\ 3^6 = 3^2·3^2·3^2 = 9·9·9 = 729 \\ {10}^3 = 10·10·10 = 1000 \\ \mathrm{จะได้} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{1024}^{100}, {729}^{100}, {1000}^{100} \\ \mathrm{ดังนั้น} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{3}^{600}<{10}^{300}<2^{1000} \end{gathered}$$