Question Number 92010 by Rio Michael last updated on 04/May/20
$$\mathrm{if}\:\mathrm{log}_{\mathrm{6}} \mathrm{30}\:=\:{a}\:\mathrm{and}\:\mathrm{log}_{\mathrm{24}} \mathrm{15}\:=\:{b} \\ $$$$\mathrm{log}_{\mathrm{12}} \mathrm{60}\:=\:? \\ $$
Answered by jagoll last updated on 04/May/20
$$\mathrm{log}_{\mathrm{12}} \left(\mathrm{60}\right)\:=\:\frac{\mathrm{log}_{\mathrm{6}} \left(\mathrm{30}\right)+\mathrm{log}_{\mathrm{6}} \left(\mathrm{2}\right)}{\mathrm{1}+\mathrm{log}_{\mathrm{6}} \left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{a}+\mathrm{log}_{\mathrm{6}} \left(\mathrm{2}\right)}{\mathrm{1}+\mathrm{log}_{\mathrm{6}} \left(\mathrm{2}\right)}\:=\:\frac{\mathrm{a}+\left(\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2b}+\mathrm{1}}\right)}{\mathrm{1}+\left(\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2b}+\mathrm{1}}\right)}\: \\ $$$$=\:\frac{\mathrm{a}\left(\mathrm{2b}+\mathrm{1}\right)+\mathrm{a}−\mathrm{b}}{\mathrm{2b}+\mathrm{1}+\mathrm{a}−\mathrm{b}}=\frac{\mathrm{2a}+\mathrm{2ab}−\mathrm{b}}{\mathrm{a}+\mathrm{b}+\mathrm{1}}\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{log}\:_{\mathrm{24}} \left(\mathrm{15}\right)=\:\frac{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{30}\right)−\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)}{\mathrm{1}+\mathrm{2}\:\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)}\:=\:\mathrm{b} \\ $$$$\mathrm{a}−\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)=\:\mathrm{b}+\mathrm{2b}\:\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right) \\ $$$$\left(\mathrm{2b}+\mathrm{1}\right)\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)=\:\mathrm{a}−\mathrm{b} \\ $$$$\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)=\:\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2b}+\mathrm{1}} \\ $$