Question Number 56351 by otchereabdullai@gmail.com last updated on 15/Mar/19
$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{2y}+\mathrm{1}\right)+\mathrm{2}=\mathrm{log}_{\mathrm{3}} \mathrm{6y}^{\mathrm{2}} −\mathrm{log}_{\mathrm{3}} \mathrm{2y} \\ $$
Answered by $@ty@m last updated on 15/Mar/19
$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{2y}+\mathrm{1}\right)+\mathrm{log}_{\mathrm{3}} \mathrm{3}^{\mathrm{2}} =\mathrm{log}_{\mathrm{3}} \mathrm{6y}^{\mathrm{2}} −\mathrm{log}_{\mathrm{3}} \mathrm{2y} \\ $$$$\mathrm{log}_{\mathrm{3}} \left\{\mathrm{9}\left(\mathrm{2}{y}+\mathrm{1}\right)\right\}=\mathrm{log}_{\mathrm{3}} \frac{\mathrm{6}{y}^{\mathrm{2}} }{\mathrm{2}{y}} \\ $$$$\mathrm{9}\left(\mathrm{2}{y}+\mathrm{1}\right)=\mathrm{3}{y} \\ $$$$\mathrm{18}{y}+\mathrm{9}=\mathrm{3}{y} \\ $$$$\mathrm{15}{y}=−\mathrm{9} \\ $$$${y}=−\frac{\mathrm{3}}{\mathrm{5}} \\ $$
Commented by otchereabdullai@gmail.com last updated on 15/Mar/19
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}! \\ $$