log_3 2, log_6 2, log


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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=((log_3 2)/(log_3 6)) , log_(12) 2=((log_3 2)/(log_3 12)) i\ ((log_6 2)/(log_3 2))=((log_3 2)/(log_3 6))×(1/(log_3 2))=(1/(log_3 6))=log_6 3 ii\((log_(12) 2)/(log_6 2))=((log_3 2)/(log_3 12))×((log_3 6)/(log_3 2))=((log_3 6)/(log_3 12))=log_(12) 6 i≠ii ⇒ 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.$
	$\log_{6} 2 = \frac{\log_{3} 2}{\log_{3} 6}, \log_{12} 2 = \frac{\log_{3} 2}{\log_{3} 12}$ $i ∖ \frac{\log_{6} 2}{\log_{3} 2} = \frac{\log_{3} 2}{\log_{3} 6} \times \frac{1}{\log_{3} 2} = \frac{1}{\log_{3} 6} = \log_{6} 3$ $ii ∖ \frac{\log_{12} 2}{\log_{6} 2} = \frac{\log_{3} 2}{\log_{3} 12} \times \frac{\log_{3} 6}{\log_{3} 2} = \frac{\log_{3} 6}{\log_{3} 12} = \log_{12} 6$ $i \neq ii \Rightarrow Not a Geometric progression$ $A similar test can be done to verify if {it}^{'} s an AP$ $if not then it must be a special Series .$
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