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Question Number 56351 by otchereabdullai@gmail.com last updated on 15/Mar/19

$${\log }_3(2\mathrm{y}+1)+2={\log }_3{6\mathrm{y}}^2-{\log }_{3}2\mathrm{y}$$

Answered by $@ty@m last updated on 15/Mar/19

$${\log }_3(2\mathrm{y}+1)+{\log }_3{3}^2={\log }_3{6\mathrm{y}}^2-{\log }_{3}2\mathrm{y}$$$${\log }_3\left\{9(2y+1)\right\}={\log }_3\frac{6y^2}{2y}$$$$9(2y+1)=3y$$$$18y+9=3y$$$$15y=-9$$$$y=-\frac{3}{5}$$

Commented by otchereabdullai@gmail.com last updated on 15/Mar/19

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}!$$

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