Tentukan nilai logaritma berikut. (3)log27+(3)log6−(3)log2

Hallo Silvyani, kakak bantu jawab ya ... Ingat kembali maka: ^a log (bxc) = ^a log b + ^a log c ^a log (b/c) = ^a log b - ^a log c ^a log (b^m) = m. ^a log b ^a log a = 1 gunakan sifat tersebut maka: ^(3)log27+^(3)log6− ^(3)log2 = ^3 log (27x6) - ^3 log 2 ------------------------------------->= ^3 log 162 - ^3 log 2 ------------------------------------->= ^3 log (162/2) ------------------------------------->= ^3 log 81 -------------------------------------> = ^3 log (3^4) -------------------------------------> = 4. ^3 log 3 -------------------------------------> = 4.1 -------------------------------------> = 4 Semoga membantu ^^