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

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Question Number 65931 by gunawan last updated on 06/Aug/19

$$\frac{{\left({\log }_{6}36\right)}^2-{\left({\log }_{3}4\right)}^2}{{\log }_3\left(\sqrt{12}\right)}=\dots$$

Answered by $@ty@m123 last updated on 06/Aug/19

$$=\frac{{\left(\log {}_6{6}^2\right)}^2-{\left(\log {}_{3}4\right)}^2}{\frac{1}{2}\log {}_{3}12}$$$$=\frac{4-{\left(\log {}_{3}4\right)}^2}{\frac{1}{2}\left\{\log {}_3\left(4\times 3\right)\right\}}$$$$=\frac{8-2{\left(\log {}_{3}4\right)}^2}{\log {}_{3}4+1}$$

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