3log6+1per 2log3- 1per 4log3

jawab

³log 6 + 1/ (²log 3) - 1/(⁴log 3) = 

= ³log 6 + ³log 2 - ³log 4
= ³log { 6(2) /4 }
= ³log 3
= 1

[tex] log_{3}(6) + \frac{1}{ log_{2}(3) } - \frac{1}{ log_{4}(3) } \\ = log_{3}(6) + log_{3}(2) - log_{3}(4) \\ = log_{3}(6 \times 2 \div 4) \\ = log_{3}(3) \\ = 1[/tex]