Question

1. 2log32=
2. 3log6+3log1/2+5log125=
3. 12log3+12log4=
4. 3log27+2log2-4log64=
5. 2log7x7log14-2log7=

Answer 1

1. 2log32 = 2log2⁵ = 5.
2. 3log6 + 3log(1/2) + 5log125
Ingat sifat logaritma: alogb + alogc = alog(bxc).
=> 3log6 + 3log(1/2) = 3log(6x1/2) = 3log3 = 1.
Jadi, 3log6 + 3log(1/2) + 5log125 = 1 + 5log5³ = 1 + 3 = 4.
3. 12log3 + 12log4 = 12log(3 x 4) = 12log12 = 1
4. 3log27 + 2log2 - 4log64
=> 3log3³ + 2log2 - 4log4³ = 3 + 1 - 3 = 1.
5. 2log7*7log14 - 2log7
=> 2log14 - 2log7
=> 2log(14:7) = 2log2 = 1