Question

а) 5 log5 25+ log 7 1^49
б)2 log 2 6- log2 9
Помогииитеее

Answer 1

$displaystyle 5log_55^2+ log_71^{49}=10log_55+ log_71=10+0=10;$

2log₂6- log₂9=log₂36- log₂9=log₂(36:9)=log₂4=2;