Loga(bc) = log a b + log a c; loga(b/c) = log a b - log a c; на каждую формулу 5...

Решите задачу:

$log_{25}125 + log_{25}5=log_{25}625=2\\\\ log_{20}80 + log_{20}100 = log_{20}8000 = 3\\\\ log_{15}3 + log_{15}5=log_{15}15=1\\\\ log_612+log_618=log_6216=3\\\\ log_832+log_82=log_864=2$

$log_7182 - log_726=log_77=1\\\\ log_{13}182-log_{13}14=log_{13}13=1\\\\ log_51200-log_548=log_525=2\\\\ log_943011-log_959=log_9729=3\\\\ log_{17}10115-log_{17}35=log_{17}289=2$

Loga(bc) = log a b + log a c; loga(b/c) = log a b - log a c; ** каждую формулу 5...