1. The problem is to evaluate the expression $\log_2 7 - \log_{14} 2$. 2. Recall the logarithm subtraction rule: $\log_a b - \log_a c = \log_a \frac{b}{c}$, but here the bases differ, so we need to convert to a common base. 3. Use the change of base formula: $\log_a b = \frac{\log b}{\log a}$ for any positive base (commonly base 10 or $e$). 4. Rewrite each term: $$\log_2 7 = \frac{\log 7}{\log 2}$$ $$\log_{14} 2 = \frac{\log 2}{\log 14}$$ 5. Substitute back: $$\log_2 7 - \log_{14} 2 = \frac{\log 7}{\log 2} - \frac{\log 2}{\log 14}$$ 6. Find a common denominator $\log 2 \cdot \log 14$: $$= \frac{\log 7 \cdot \log 14}{\log 2 \cdot \log 14} - \frac{\log 2 \cdot \log 2}{\log 14 \cdot \log 2} = \frac{\log 7 \cdot \log 14 - (\log 2)^2}{\log 2 \cdot \log 14}$$ 7. Note that $\log 14 = \log (2 \times 7) = \log 2 + \log 7$. 8. Substitute $\log 14$: $$= \frac{\log 7 (\log 2 + \log 7) - (\log 2)^2}{\log 2 (\log 2 + \log 7)} = \frac{\log 7 \cdot \log 2 + (\log 7)^2 - (\log 2)^2}{\log 2 (\log 2 + \log 7)}$$ 9. The numerator is a difference of squares rearranged: $$= \frac{(\log 7)^2 + \log 7 \cdot \log 2 - (\log 2)^2}{\log 2 (\log 2 + \log 7)}$$ 10. This expression is simplified as is; numerical approximation can be done if needed. Final answer: $$\log_2 7 - \log_{14} 2 = \frac{(\log 7)^2 + \log 7 \cdot \log 2 - (\log 2)^2}{\log 2 (\log 2 + \log 7)}$$