a)log6(12)+log6(3)=б)log49(84)-log49(12)=в)[tex] - вопрос №61263498449129310191477 от ange4 20.08.2022 19:12

Question

Математика: a)log6(12)+log6(3)=б)log49(84)-log49(12)=в)[tex] rac{lg9}{lg3} = [/tex]г)log2log2(2)д)2^1-log2(7)​

ange4 · Answer

a)log6(12)+log6(3) = log6(36) = 2б)log49(84)-log49(12) = log49(7) = 1/2в)lg9/lg3 = log3(9) = 2г)log2log2(2) = log2(1) = 0д)2^1-log2(7) = 2/2^log2(7) = 2/7...

a)log6(12)+log6(3)=б)log49(84)-log49(12)=в)г)log2log2(2)д)2^1-log2(7)​

a)log6(12)+log6(3)=
б)log49(84)-log49(12)=
в)
$\frac{lg9}{lg3} =$
г)log2log2(2)
д)2^1-log2(7)