Question Number 147334 by Tawa11 last updated on 19/Jul/21
$$\mathrm{If}\:\:\:\:\mathrm{log}_{\mathrm{6}} \mathrm{30}\:\:\:=\:\:\:\mathrm{a},\:\:\:\:\:\:\:\:\:\mathrm{log}_{\mathrm{15}} \mathrm{24}\:\:\:=\:\:\:\mathrm{b},\:\:\:\:\:\:\:\:\:\:\:\mathrm{evaluate}:\:\:\:\mathrm{log}_{\mathrm{12}} \mathrm{60} \\ $$
Commented by otchereabdullai@gmail.com last updated on 20/Jul/21
$$\mathrm{nice}! \\ $$
Answered by bramlexs22 last updated on 20/Jul/21
$$\:\:\mathrm{If}\:\begin{cases}{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{30}\right)=\mathrm{a}}\\{\mathrm{log}\:_{\mathrm{15}} \left(\mathrm{24}\right)=\mathrm{b}}\end{cases}\:\mathrm{Evaluate}\:\mathrm{log}\:_{\mathrm{12}} \left(\mathrm{60}\right). \\ $$$$\equiv\:\mathrm{solution}\:\equiv \\ $$$$\left(\bullet\right)\:\mathrm{a}=\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{6}×\mathrm{5}\right)=\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{5}\right) \\ $$$$\:\Rightarrow\:\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{5}\right)=\mathrm{a}−\mathrm{1} \\ $$$$\left(\bullet\right)\mathrm{b}=\:\mathrm{log}_{\mathrm{15}} \left(\mathrm{24}\right)=\frac{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{24}\right)}{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{15}\right)}=\frac{\mathrm{1}+\mathrm{2log}_{\mathrm{6}} \left(\mathrm{2}\right)\:}{\mathrm{a}−\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{3}\right)}\: \\ $$$$\Rightarrow\mathrm{ab}−\mathrm{b}+\mathrm{b}\left(\mathrm{1}−\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)\right)=\mathrm{1}+\mathrm{2log}\:_{\mathrm{6}} \left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{ab}−\mathrm{1}=\left(\mathrm{2}+\mathrm{b}\right)\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)=\frac{\mathrm{ab}−\mathrm{1}}{\mathrm{b}+\mathrm{2}} \\ $$$$\left(\bullet\right)\mathrm{log}\:_{\mathrm{12}} \left(\mathrm{60}\right)=\frac{\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{5}\right)+\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)}{\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)} \\ $$$$=\:\frac{\mathrm{1}+\mathrm{a}−\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)}{\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)}=\frac{\mathrm{a}+\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)}{\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{a}+\frac{\mathrm{ab}−\mathrm{1}}{\mathrm{b}+\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{ab}−\mathrm{1}}{\mathrm{b}+\mathrm{2}}}\:=\:\frac{\mathrm{2ab}+\mathrm{2a}−\mathrm{1}}{\mathrm{ab}+\mathrm{b}+\mathrm{1}}\:. \\ $$
Commented by Tawa11 last updated on 20/Jul/21
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$