x-2-log-3-x-2-2x-2-3-log-9-2x-3-3log-3-x-2x-3-

Question Number 204300 by depressiveshrek last updated on 11/Feb/24

x^{2} \log_{3} x^{2} - (2 x^{2} + 3) \log_{9} (2 x + 3) = {3 \log}_{3} (\frac{x}{2 x + 3})

Answered by Rasheed.Sindhi last updated on 11/Feb/24

x^{2} \log_{3} x^{2} - (2 x^{2} + 3) \log_{9} (2 x + 3) = {3 \log}_{3} (\frac{x}{2 x + 3})

\log_{9} (2 x + 3) = \frac{\log_{3} (2 x + 3)}{\log_{3} 9} = \frac{\log_{3} (2 x + 3)}{2}

2 x^{2} \log_{3} x - (2 x^{2} + 3) (\frac{\log_{3} (2 x + 3)}{2}) = {3 \log}_{3} (\frac{x}{2 x + 3})

2 x^{2} \log_{3} x - (2 x^{2} + 3) \log_{3} (2 x + 3) = {6 \log}_{3} (\frac{x}{2 x + 3})

4 x^{2} \log_{3} x - (2 x^{2} + 3) \log_{3} (2 x + 3) = \log_{3} {(\frac{x}{2 x + 3})}^{6}

\log_{3} x^{4 x^{2}} - \log_{3} {(2 x + 3)}^{2 x^{2} + 3} = \log_{3} {(\frac{x}{2 x + 3})}^{6}

\log_{3} (x^{4 x^{2}} \div {(2 x + 3)}^{2 x^{2} + 3}) = \log_{3} {(\frac{x}{2 x + 3})}^{6}

x^{4 x^{2}} \div {(2 x + 3)}^{2 x^{2} + 3} = {(\frac{x}{2 x + 3})}^{6}

\frac{x^{4 x^{2}}}{{(2 x + 3)}^{2 x^{2} + 3}} = {(\frac{x}{2 x + 3})}^{6}

Commented by AST last updated on 12/Feb/24

\Rightarrow x^{4 x^{2} - 6} = {(2 x + 3)}^{2 x^{2} - 3} \Rightarrow {(x^{2})}^{2 x^{2} - 3} = {(2 x + 3)}^{2 x^{2} - 3}

\Rightarrow {(\frac{x^{2}}{2 x + 3})}^{2 x^{2} - 3} = 1 \Rightarrow \frac{x^{2}}{2 x + 3} = 1 \lor 2 x^{2} - 3 = 0

\Rightarrow x = 3, x = \frac{\sqrt{6}}{2},

Commented by Rasheed.Sindhi last updated on 12/Feb/24

T h a n k s s i r!