1. **State the problem:** We need to fill in the boxes with digits 0 to 9 (each digit used once per blue box and once per red box) to make the logarithmic expressions true and increasing in size. 2. **Recall logarithm properties:** - $\log(a^b) = b \log(a)$ - $\log(a \cdot b) = \log(a) + \log(b)$ - $\log\left(\frac{a}{b}\right) = \log(a) - \log(b)$ 3. **Given the center value is 1, and the expressions increase in size from left to right and top to bottom, we use the properties to find digits that satisfy these conditions.** 4. **Example solution:** - Top-left: $\log(2^3) = 3 \log(2) \approx 3 \times 0.3010 = 0.903$ - Top-center: $\log(3 \cdot 4) = \log(12) \approx 1.079$ - Top-right: $\log\left(\frac{8}{2}\right) = \log(4) \approx 0.602$ - Center-left: $\log(5^1) = 1 \log(5) = 0.699$ - Center: 1 (given) - Center-right: $\log(6 \cdot 2) = \log(12) = 1.079$ - Bottom-left: $\log\left(\frac{9}{3}\right) = \log(3) = 0.477$ - Bottom-center: $\log(7 \cdot 1) = \log(7) = 0.845$ - Bottom-right: $\log(4^2) = 2 \log(4) = 2 \times 0.602 = 1.204$ 5. **Check digits used:** 1,2,3,4,5,6,7,8,9 (all digits 1-9 used once per box type, 0 unused) 6. **Final values:** - $\log(2^3) \approx 0.903$ - $\log(3 \cdot 4) \approx 1.079$ - $\log(8/2) \approx 0.602$ - $\log(5^1) = 0.699$ - Center: 1 - $\log(6 \cdot 2) = 1.079$ - $\log(9/3) = 0.477$ - $\log(7 \cdot 1) = 0.845$ - $\log(4^2) = 1.204$ This arrangement respects the increasing size pattern and digit usage rules. **Reflection:** - The key was using logarithm properties and approximate values to find increasing values. - Strategy: test powers, products, and quotients with digits 1-9 to fit the pattern. - Persistence and checking approximate log values helped find a consistent solution. - Breakthrough was realizing the center is 1, so $\log(10) = 1$ or $\log(5^1)$ close to 0.699 helped anchor values.