ini gimana caranya...3log1/27+4log40+4log32/5+2log⅛... yg bisa jawab aku kasih poin lumayan

Poin lumayan...
Hmm... gmn dg ILMU yg kmu terima ;)

(a^n)'log (b^m) = (m/n) . a'log b

•••

³log (1/27) + ⁴log 40 + ⁴log (32/5) + ²log (1/8)

= ³log (3^-3) + ⁴log (40 × 32/5) + ²log (2^-3)

= (-3) . ³log 3 + (2^2)'log (2^3 × 2^5) + (-3) . ²log 2

= -3 + (8/2) . ²log 2 + (-3)

= -6 + 4

= -2