Question Number 204300 by depressiveshrek last updated on 11/Feb/24
$${x}^{\mathrm{2}} \mathrm{log}_{\mathrm{3}} {x}^{\mathrm{2}} −\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\mathrm{log}_{\mathrm{9}} \left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{3log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 11/Feb/24
$${x}^{\mathrm{2}} \mathrm{log}_{\mathrm{3}} {x}^{\mathrm{2}} −\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\mathrm{log}_{\mathrm{9}} \left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{3log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{log}_{\mathrm{9}} \left(\mathrm{2}{x}+\mathrm{3}\right)=\frac{\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{3}\right)\:}{\mathrm{log}_{\mathrm{3}} \mathrm{9}\:}=\frac{\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{3}\right)\:}{\mathrm{2}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \mathrm{log}_{\mathrm{3}} {x}−\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\left(\frac{\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{3}\right)\:}{\mathrm{2}\:}\right)=\mathrm{3log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)\:\:\: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \mathrm{log}_{\mathrm{3}} {x}−\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{6log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)\:\:\: \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \mathrm{log}_{\mathrm{3}} {x}−\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{6}} \:\:\: \\ $$$$\mathrm{log}_{\mathrm{3}} {x}^{\mathrm{4}{x}^{\mathrm{2}} } −\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}} =\mathrm{log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{6}} \:\: \\ $$$$\mathrm{log}_{\mathrm{3}} \left({x}^{\mathrm{4}{x}^{\mathrm{2}} } \boldsymbol{\div}\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}} \right)=\mathrm{log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{6}} \\ $$$${x}^{\mathrm{4}{x}^{\mathrm{2}} } \boldsymbol{\div}\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}} =\left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{6}} \\ $$$$\frac{{x}^{\mathrm{4}{x}^{\mathrm{2}} } }{\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}} }=\left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{6}} \\ $$
Commented by AST last updated on 12/Feb/24
$$\Rightarrow{x}^{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{6}} =\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}} \Rightarrow\left({x}^{\mathrm{2}} \right)^{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}} =\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}} \\ $$$$\Rightarrow\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}} =\mathrm{1}\Rightarrow\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}+\mathrm{3}}=\mathrm{1}\vee\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3},{x}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}, \\ $$
Commented by Rasheed.Sindhi last updated on 12/Feb/24
$$\mathcal{T}{hanks}\:{sir}! \\ $$