[แมวสามสี]

ข้อ 3 ครับ

$ln(log_2 3)-[ln(log_4 3)+...+ln(log_n (n-1))]$=$\frac{1}{2}ln36$

$ln(log_2 3)-[ln(\frac{log3}{log4}\bullet \frac{log4}{log5}\bullet ...\bullet\frac{log(n-1)}{log n})] = ln6$

$ln(\frac{log3}{log2})-ln(\frac{log3}{log n}) = ln6$

$ln(\frac{log3}{log2}\bullet \frac{log n}{log3}) = ln6$

$log_2 n = 6 $

$n = 2^6$