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Question Number 92010 by Rio Michael last updated on 04/May/20

if log_6 30 = a and log_(24) 15 = b log_(12) 60 = ?

$$\mathrm{if}\:\mathrm{log}_{\mathrm{6}} \mathrm{30}\:=\:{a}\:\mathrm{and}\:\mathrm{log}_{\mathrm{24}} \mathrm{15}\:=\:{b} \\ $$$$\mathrm{log}_{\mathrm{12}} \mathrm{60}\:=\:? \\ $$

Answered by jagoll last updated on 04/May/20

log_(12) (60) = ((log_6 (30)+log_6 (2))/(1+log_6 (2))) =((a+log_6 (2))/(1+log_6 (2))) = ((a+(((a−b)/(2b+1))))/(1+(((a−b)/(2b+1))))) = ((a(2b+1)+a−b)/(2b+1+a−b))=((2a+2ab−b)/(a+b+1)) (1) log _(24) (15)= ((log _6 (30)−log _6 (2))/(1+2 log _6 (2))) = b a−log _6 (2)= b+2b log _6 (2) (2b+1)log _6 (2)= a−b log _6 (2)= ((a−b)/(2b+1))

$$\mathrm{log}_{\mathrm{12}} \left(\mathrm{60}\right)\:=\:\frac{\mathrm{log}_{\mathrm{6}} \left(\mathrm{30}\right)+\mathrm{log}_{\mathrm{6}} \left(\mathrm{2}\right)}{\mathrm{1}+\mathrm{log}_{\mathrm{6}} \left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{a}+\mathrm{log}_{\mathrm{6}} \left(\mathrm{2}\right)}{\mathrm{1}+\mathrm{log}_{\mathrm{6}} \left(\mathrm{2}\right)}\:=\:\frac{\mathrm{a}+\left(\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2b}+\mathrm{1}}\right)}{\mathrm{1}+\left(\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2b}+\mathrm{1}}\right)}\: \\ $$$$=\:\frac{\mathrm{a}\left(\mathrm{2b}+\mathrm{1}\right)+\mathrm{a}−\mathrm{b}}{\mathrm{2b}+\mathrm{1}+\mathrm{a}−\mathrm{b}}=\frac{\mathrm{2a}+\mathrm{2ab}−\mathrm{b}}{\mathrm{a}+\mathrm{b}+\mathrm{1}}\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{log}\:_{\mathrm{24}} \left(\mathrm{15}\right)=\:\frac{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{30}\right)−\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)}{\mathrm{1}+\mathrm{2}\:\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)}\:=\:\mathrm{b} \\ $$$$\mathrm{a}−\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)=\:\mathrm{b}+\mathrm{2b}\:\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right) \\ $$$$\left(\mathrm{2b}+\mathrm{1}\right)\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)=\:\mathrm{a}−\mathrm{b} \\ $$$$\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)=\:\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2b}+\mathrm{1}} \\ $$

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