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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

$$\mathrm{log}_{\mathrm{3}} \mathrm{2},\:\:\mathrm{log}_{\mathrm{6}} \mathrm{2},\:\:\mathrm{log}_{\mathrm{12}} \mathrm{2}\:\:\:\mathrm{are}\:\mathrm{in} \\ $$

Answered by 4635 last updated on 20/Jun/20

.log _3 2=((ln 2)/(ln 3)) ,.log _6 2=((ln 2)/(ln 2+ln 3)) .log _(12) 2=((ln 2)/(2ln 2+ln 3))

$$.\mathrm{log}\:_{\mathrm{3}} \mathrm{2}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}\:,.\mathrm{log}\:_{\mathrm{6}} \mathrm{2}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}} \\ $$$$.\mathrm{log}\:_{\mathrm{12}} \mathrm{2}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}} \\ $$

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.

$$\mathrm{log}_{\mathrm{6}} \mathrm{2}=\frac{\mathrm{log}_{\mathrm{3}} \mathrm{2}}{\mathrm{log}_{\mathrm{3}} \mathrm{6}}\:\:,\:\mathrm{log}_{\mathrm{12}} \mathrm{2}=\frac{\mathrm{log}_{\mathrm{3}} \mathrm{2}}{\mathrm{log}_{\mathrm{3}} \mathrm{12}} \\ $$$$\mathrm{i}\backslash\:\frac{\mathrm{log}_{\mathrm{6}} \mathrm{2}}{\mathrm{log}_{\mathrm{3}} \mathrm{2}}=\frac{\mathrm{log}_{\mathrm{3}} \mathrm{2}}{\mathrm{log}_{\mathrm{3}} \mathrm{6}}×\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{3}} \mathrm{2}}=\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{3}} \mathrm{6}}=\mathrm{log}_{\mathrm{6}} \mathrm{3} \\ $$$$\mathrm{ii}\backslash\frac{\mathrm{log}_{\mathrm{12}} \mathrm{2}}{\mathrm{log}_{\mathrm{6}} \mathrm{2}}=\frac{\mathrm{log}_{\mathrm{3}} \mathrm{2}}{\mathrm{log}_{\mathrm{3}} \mathrm{12}}×\frac{\mathrm{log}_{\mathrm{3}} \mathrm{6}}{\mathrm{log}_{\mathrm{3}} \mathrm{2}}=\frac{\mathrm{log}_{\mathrm{3}} \mathrm{6}}{\mathrm{log}_{\mathrm{3}} \mathrm{12}}=\mathrm{log}_{\mathrm{12}} \mathrm{6} \\ $$$$\mathrm{i}\neq\mathrm{ii}\:\:\Rightarrow\:\mathrm{Not}\:\mathrm{a}\:\mathrm{Geometric}\:\mathrm{progression} \\ $$$$\mathcal{A}\:\mathrm{similar}\:\mathrm{test}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{to}\:\mathrm{verify}\:\mathrm{if}\:\mathrm{it}'\mathrm{s}\:\mathrm{an}\:\mathrm{AP} \\ $$$$\mathrm{if}\:\mathrm{not}\:\mathrm{then}\:\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:\mathrm{a}\:\mathrm{special}\:\mathrm{Series}. \\ $$

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