Question Number 198413 by mokys last updated on 19/Oct/23
Commented by mokys last updated on 19/Oct/23
$${find}\:{log}_{\mathrm{12}} \mathrm{60} \\ $$
Answered by cortano12 last updated on 19/Oct/23
$$\:\:\begin{cases}{\mathrm{log}\:_{\mathrm{15}} \left(\mathrm{24}\right)=\mathrm{b}\Rightarrow\mathrm{log}\:_{\mathrm{24}} \left(\mathrm{15}\right)=\frac{\mathrm{1}}{\mathrm{b}}}\\{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{30}\right)=\mathrm{a}\Rightarrow\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{5}\right)=\mathrm{a}}\end{cases} \\ $$$$\:\:\frac{\mathrm{log}\:_{\mathrm{6}} \mathrm{15}}{\mathrm{log}\:_{\mathrm{6}} \mathrm{24}}\:=\frac{\mathrm{1}}{\mathrm{b}}\:\Rightarrow\frac{\mathrm{a}−\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \mathrm{3}}{\mathrm{1}+\mathrm{2log}\:_{\mathrm{6}} \mathrm{2}}=\frac{\mathrm{1}}{\mathrm{b}} \\ $$$$\:\:\frac{\mathrm{a}−\mathrm{log}\:_{\mathrm{6}} \mathrm{2}}{\mathrm{1}+\mathrm{2log}\:_{\mathrm{6}} \mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{b}}\:\Rightarrow\mathrm{log}\:_{\mathrm{6}} \mathrm{2}\:=\:\frac{\mathrm{ab}−\mathrm{1}}{\mathrm{b}+\mathrm{2}} \\ $$$$\: \\ $$$$\:\:\mathrm{log}\:_{\mathrm{12}} \mathrm{60}\:=\:\frac{\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \mathrm{5}+\mathrm{log}\:_{\mathrm{6}} \mathrm{2}}{\mathrm{1}+\mathrm{log}\:_{\mathrm{6}} \mathrm{2}} \\ $$$$\:\:=\:\frac{\mathrm{1}+\mathrm{a}−\mathrm{1}+\frac{\mathrm{ab}−\mathrm{1}}{\mathrm{b}+\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{ab}−\mathrm{1}}{\mathrm{b}+\mathrm{2}}} \\ $$$$\:=\:\frac{\mathrm{2ab}+\mathrm{2a}−\mathrm{1}}{\mathrm{ab}+\mathrm{b}+\mathrm{1}}\: \\ $$$$ \\ $$$$\: \\ $$