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log-6-36-2-log-3-4-2-log-3-12-




Question Number 65931 by gunawan last updated on 06/Aug/19
(((log_6 36)^2 −(log_3 4)^2 )/(log_3 ((√(12)))))=...
$$\frac{\left(\mathrm{log}_{\mathrm{6}} \mathrm{36}\right)^{\mathrm{2}} −\left(\mathrm{log}_{\mathrm{3}} \mathrm{4}\right)^{\mathrm{2}} }{\mathrm{log}_{\mathrm{3}} \left(\sqrt{\mathrm{12}}\right)}=… \\ $$
Answered by $@ty@m123 last updated on 06/Aug/19
=(((log _6 6^2 )^2 −(log _3 4)^2 )/((1/2)log _3 12))  =((4−(log _3 4)^2 )/((1/2){log _3 (4×3)}))  =((8−2(log _3 4)^2 )/(log _3 4+1))
$$=\frac{\left(\mathrm{log}\:_{\mathrm{6}} \mathrm{6}^{\mathrm{2}} \right)^{\mathrm{2}} −\left(\mathrm{log}\:_{\mathrm{3}} \mathrm{4}\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:_{\mathrm{3}} \mathrm{12}} \\ $$$$=\frac{\mathrm{4}−\left(\mathrm{log}\:_{\mathrm{3}} \mathrm{4}\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}×\mathrm{3}\right)\right\}} \\ $$$$=\frac{\mathrm{8}−\mathrm{2}\left(\mathrm{log}\:_{\mathrm{3}} \mathrm{4}\right)^{\mathrm{2}} }{\mathrm{log}\:_{\mathrm{3}} \mathrm{4}+\mathrm{1}} \\ $$