Question Number 65931 by gunawan last updated on 06/Aug/19
$$\frac{\left(\mathrm{log}_{\mathrm{6}} \mathrm{36}\right)^{\mathrm{2}} −\left(\mathrm{log}_{\mathrm{3}} \mathrm{4}\right)^{\mathrm{2}} }{\mathrm{log}_{\mathrm{3}} \left(\sqrt{\mathrm{12}}\right)}=... \\ $$
Answered by $@ty@m123 last updated on 06/Aug/19
$$=\frac{\left(\mathrm{log}\:_{\mathrm{6}} \mathrm{6}^{\mathrm{2}} \right)^{\mathrm{2}} −\left(\mathrm{log}\:_{\mathrm{3}} \mathrm{4}\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:_{\mathrm{3}} \mathrm{12}} \\ $$$$=\frac{\mathrm{4}−\left(\mathrm{log}\:_{\mathrm{3}} \mathrm{4}\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}×\mathrm{3}\right)\right\}} \\ $$$$=\frac{\mathrm{8}−\mathrm{2}\left(\mathrm{log}\:_{\mathrm{3}} \mathrm{4}\right)^{\mathrm{2}} }{\mathrm{log}\:_{\mathrm{3}} \mathrm{4}+\mathrm{1}} \\ $$