1. We need to evaluate the expression $$\frac{2 + \log_2 3}{1 + \log_2 3} + \frac{3 + \log_3 4}{1 + \log_3 2}$$. 2. First, let's analyze the first fraction: $$\frac{2 + \log_2 3}{1 + \log_2 3}$$. We can rewrite the numerator as $$1 + 1 + \log_2 3$$ to see if it helps. 3. Notice that $$\log_2 3$$ and $$1$$ are common in numerator and denominator: Numerator: $$2 + \log_2 3 = 1 + (1 + \log_2 3)$$ Denominator: $$1 + \log_2 3$$. Therefore, the fraction becomes: $$\frac{1 + (1 + \log_2 3)}{1 + \log_2 3} = 1 + \frac{1}{1 + \log_2 3}$$. 4. Now for the second fraction: $$\frac{3 + \log_3 4}{1 + \log_3 2}$$. Rewrite $$3$$ as $$1 + 2$$: $$3 + \log_3 4 = 1 + 2 + \log_3 4$$. Group terms: $$1 + (2 + \log_3 4)$$. 5. Notice that: $$\log_3 4 = \log_3 (2^2) = 2 \log_3 2$$. Therefore: $$2 + \log_3 4 = 2 + 2 \log_3 2 = 2(1 + \log_3 2)$$. 6. Substitute back: $$\frac{3 + \log_3 4}{1 + \log_3 2} = \frac{1 + 2(1 + \log_3 2)}{1 + \log_3 2} = \frac{1 + 2 \cdot (1 + \log_3 2)}{1 + \log_3 2}$$. Split the fraction: $$= \frac{1}{1 + \log_3 2} + \frac{2 (1 + \log_3 2)}{1 + \log_3 2} = \frac{1}{1 + \log_3 2} + 2$$. 7. Now the original expression is: $$1 + \frac{1}{1 + \log_2 3} + \frac{1}{1 + \log_3 2} + 2 = 3 + \frac{1}{1 + \log_2 3} + \frac{1}{1 + \log_3 2}$$. 8. Use the property of logarithms base change: $$\log_3 2 = \frac{1}{\log_2 3}$$. Define $$x = \log_2 3$$ so: $$1 + \log_3 2 = 1 + \frac{1}{x} = \frac{x + 1}{x}$$. Similarly: $$\frac{1}{1 + \log_3 2} = \frac{1}{\frac{x + 1}{x}} = \frac{x}{x + 1}$$. And: $$\frac{1}{1 + \log_2 3} = \frac{1}{1 + x}$$. 9. Substitute back: $$3 + \frac{1}{1 + x} + \frac{x}{x + 1} = 3 + \frac{1}{1 + x} + \frac{x}{1 + x} = 3 + \frac{1 + x}{1 + x} = 3 + 1 = 4$$. **Final answer:** $$4$$