log-3-2y-1-2-log-3-6y-2-log-3-2y-

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

\log_{3} (2 y + 1) + 2 = \log_{3} {6 y}^{2} - \log_{3} 2 y

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

\log_{3} (2 y + 1) + \log_{3} 3^{2} = \log_{3} {6 y}^{2} - \log_{3} 2 y

\log_{3} {9 (2 y + 1)} = \log_{3} \frac{6 y^{2}}{2 y}

9 (2 y + 1) = 3 y

18 y + 9 = 3 y

15 y = - 9

y = - \frac{3}{5}

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

Thank you sir!