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 $
$n = 2^6$
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