Question

Sa se arate ca numarul n= log3 2/1 -log3 2/3 + log3 4/3- log3 4/5+log3 6/5 -log3 6/7+ log3 8/7- log3 8/9 este numar natural.

Answer 1

n=log3(2/1*3/2)+log3(4/3*5/4)+log3(6/5*7/6)+log3(8/7*9/8)
n=log3 3+ log3 5/3+ log3 7/5+ log3 9/7
n=log3(3*5/3*7/5*9/7)
n=log3 9 (se reduc toate nr inafara de 9)
n=2 deci n este natural

Answer 2

$\displaystyle log_3 \frac{2}{1} -log_3 \frac{2}{3} +log_3 \frac{4}{3} -log_3 \frac{4}{5} +log_3 \frac{6}{5} -log_3 \frac{6}{7} +log_3 \frac{8}{7} -log_3 \frac{8}{9} =$

[tex]\displaystyle =log_3 \left( \frac{2}{1} : \frac{2}{3} \cdot \frac{4}{3} : \frac{4}{5} \cdot \frac{6}{5} : \frac{6}{7} \cdot \frac{8}{7} : \frac{8}{9} \right)= [/tex]

[tex]\displaystyle log_3\left( \frac{\not2}{1} \cdot \frac{\not3}{\not2} \cdot \frac{\not4}{\not3} \cdot \frac{\not5}{\not4} \cdot \frac{\not6}{\not5} \cdot \frac{\not7}{\not6} \cdot \frac{\not8}{\not7} \cdot \frac{9}{\not8} \right) =log_39=log_33^2=2log_33=2 \in N[/tex]