Question

4 va rog …………………………….

Answer 1

$\displaystyle A=log_3\frac{2}{1} +log_3\frac{3}{2} +log_3\frac{4}{3} +...+log_3\frac{9}{8} =log_3\Bigg(\frac{\not2}{1} \cdot \frac{\not3}{\not2} \cdot \frac{\not4}{\not3} \cdot ...\cdot\frac{\not8}{\not7} \cdot \frac{9}{\not8} \Bigg)=\\\\=log_39=log_33^2=2log_33=2 \cdot 1=2\in \mathbb{N}$