Question Number 211706 by Nadirhashim last updated on 17/Sep/24
$$\:\:\:\:\:\boldsymbol{{if}}\:\:\boldsymbol{{x}}^{\boldsymbol{{log}}\mathrm{27}} +\:\:\:\mathrm{9}^{\boldsymbol{{logx}}} =\mathrm{36}\:\boldsymbol{{find}}\:\boldsymbol{{x}} \\ $$
Answered by mehdee7396 last updated on 17/Sep/24
$${x}^{\mathrm{3}{log}\mathrm{3}} +{x}^{\mathrm{2}{log}\mathrm{3}} =\mathrm{36}\:\: \\ $$$${x}^{{log}\mathrm{3}} ={y}\Rightarrow{y}^{\mathrm{3}} +{y}^{\mathrm{2}} −\mathrm{36}=\mathrm{0} \\ $$$$\left({y}−\mathrm{3}\right)\left({y}^{\mathrm{2}} +\mathrm{4}{y}+\mathrm{12}\right)=\mathrm{0} \\ $$$${if}\:\:{x}\in\mathbb{R}\Rightarrow{y}=\mathrm{3}={x}^{{log}\mathrm{3}} \Rightarrow{x}=\mathrm{10}\:\checkmark \\ $$$$ \\ $$