Question Number 199389 by depressiveshrek last updated on 02/Nov/23
$$\mathrm{log}_{\mathrm{12}} \mathrm{60}=? \\ $$$$\mathrm{log}_{\mathrm{6}} \mathrm{30}={a} \\ $$$$\mathrm{log}_{\mathrm{15}} \mathrm{24}={b} \\ $$
Answered by cortano12 last updated on 03/Nov/23
$$\:\mathrm{log}\:_{\mathrm{12}} \left(\mathrm{60}\right)=\frac{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{6}.\mathrm{5}.\mathrm{2}\right)}{\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{6}.\mathrm{2}\right)} \\ $$$$\:\:=\:\frac{\mathrm{2ab}+\mathrm{2a}−\mathrm{1}}{\mathrm{ab}+\mathrm{b}+\mathrm{1}}\: \\ $$
Commented by depressiveshrek last updated on 03/Nov/23
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{how}\:\mathrm{you}\:\mathrm{got}\:\mathrm{there}… \\ $$
Answered by mr W last updated on 04/Nov/23
$$\mathrm{log}_{\mathrm{15}} \:\mathrm{24}=\frac{\mathrm{log}_{\mathrm{6}} \:\mathrm{24}}{\mathrm{log}_{\mathrm{6}} \:\mathrm{15}}=\frac{\mathrm{log}_{\mathrm{6}} \:\mathrm{6}+\mathrm{2log}_{\mathrm{6}} \:\mathrm{2}}{\mathrm{log}_{\mathrm{6}} \:\mathrm{30}−\mathrm{log}_{\mathrm{6}} \:\mathrm{2}} \\ $$$$=\frac{\mathrm{1}+\mathrm{2log}_{\mathrm{6}} \:\mathrm{2}}{\mathrm{log}_{\mathrm{6}} \:\mathrm{30}−\mathrm{log}_{\mathrm{6}} \:\mathrm{2}} \\ $$$$=\frac{\mathrm{1}+\mathrm{2log}_{\mathrm{6}} \:\mathrm{2}}{{a}−\mathrm{log}_{\mathrm{6}} \:\mathrm{2}}={b} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{6}} \:\mathrm{2}=\frac{{ab}−\mathrm{1}}{\mathrm{2}+{b}} \\ $$$$ \\ $$$$\mathrm{log}_{\mathrm{12}} \:\mathrm{60} \\ $$$$=\frac{\mathrm{log}_{\mathrm{6}} \:\mathrm{30}+\mathrm{log}_{\mathrm{6}} \:\mathrm{2}}{\mathrm{log}_{\mathrm{6}} \:\mathrm{6}+\mathrm{log}_{\mathrm{6}} \:\mathrm{2}} \\ $$$$=\frac{\mathrm{log}_{\mathrm{6}} \:\mathrm{30}+\mathrm{log}_{\mathrm{6}} \:\mathrm{2}}{\mathrm{1}+\mathrm{log}_{\mathrm{6}} \:\mathrm{2}} \\ $$$$=\frac{{a}+\frac{{ab}−\mathrm{1}}{\mathrm{2}+{b}}}{\mathrm{1}+\frac{{ab}−\mathrm{1}}{\mathrm{2}+{b}}} \\ $$$$=\frac{\mathrm{2}{a}\left({b}+\mathrm{1}\right)−\mathrm{1}}{\left({a}+\mathrm{1}\right){b}+\mathrm{1}}\:\checkmark \\ $$