1. **State the problem:** Sort the following values from least to greatest: Top-left: $\log_4(26)$ Top-center: $\infty$ Top-right: $\sum_{i=2}^3 i$ Center-left: $\sqrt{14}$ Center-center: $2!$ Center-right: $\int_2^7 x \, dx$ Bottom-left: $\frac{2\pi}{2}$ Bottom-center: $e^3$ Bottom-right: $\frac{12}{13}$ 2. **Evaluate each expression:** - $\log_4(26)$: Since $4^2=16$ and $4^3=64$, $\log_4(26)$ is between 2 and 3. More precisely, $\log_4(26) = \frac{\ln 26}{\ln 4} \approx \frac{3.258}{1.386} \approx 2.35$. - $\infty$: Infinity, the largest value. - $\sum_{i=2}^3 i = 2 + 3 = 5$. - $\sqrt{14} \approx 3.7417$. - $2! = 2 \times 1 = 2$. - $\int_2^7 x \, dx = \left[ \frac{x^2}{2} \right]_2^7 = \frac{7^2}{2} - \frac{2^2}{2} = \frac{49}{2} - \frac{4}{2} = \frac{45}{2} = 22.5$. - $\frac{2\pi}{2} = \pi \approx 3.1416$. - $e^3 \approx 20.0855$. - $\frac{12}{13} \approx 0.9231$. 3. **Sort the values from least to greatest:** $\frac{12}{13} \approx 0.9231 < 2! = 2 < \pi \approx 3.1416 < \sqrt{14} \approx 3.7417 < \log_4(26) \approx 2.35$ (Note: $2.35$ is less than $3.1416$, so reorder) Correct order: $\frac{12}{13} \approx 0.9231 < 2! = 2 < \log_4(26) \approx 2.35 < \pi \approx 3.1416 < \sqrt{14} \approx 3.7417 < \sum_{i=2}^3 i = 5 < e^3 \approx 20.0855 < \int_2^7 x \, dx = 22.5 < \infty$ 4. **Final answer:** $$\frac{12}{13} < 2! < \log_4(26) < \pi < \sqrt{14} < \sum_{i=2}^3 i < e^3 < \int_2^7 x \, dx < \infty$$