log-3-2-log-6-2-log-12-2-are-in-

Question Number 99352 by 659083337 last updated on 20/Jun/20

\log_{3} 2, \log_{6} 2, \log_{12} 2 are in

Answered by 4635 last updated on 20/Jun/20

. \log_{3} 2 = \frac{\ln 2}{\ln 3}, . \log_{6} 2 = \frac{\ln 2}{\ln 2 + \ln 3}

. \log_{12} 2 = \frac{\ln 2}{2 \ln 2 + \ln 3}

Commented by Ar Brandon last updated on 21/Jun/20

Il n'a pas demandé de le transformer en ln mais de chercher de quel type de suite qu'il s'agit.��

Commented by Ar Brandon last updated on 21/Jun/20

Takouchouang��

Answered by Ar Brandon last updated on 21/Jun/20

\log_{6} 2 = \frac{\log_{3} 2}{\log_{3} 6}, \log_{12} 2 = \frac{\log_{3} 2}{\log_{3} 12}

i ∖ \frac{\log_{6} 2}{\log_{3} 2} = \frac{\log_{3} 2}{\log_{3} 6} \times \frac{1}{\log_{3} 2} = \frac{1}{\log_{3} 6} = \log_{6} 3

ii ∖ \frac{\log_{12} 2}{\log_{6} 2} = \frac{\log_{3} 2}{\log_{3} 12} \times \frac{\log_{3} 6}{\log_{3} 2} = \frac{\log_{3} 6}{\log_{3} 12} = \log_{12} 6

i \neq ii \Rightarrow Not a Geometric progression

A similar test can be done to verify if {it}^{'} s an AP

if not then it must be a special Series .