Question Number 195087 by bett last updated on 23/Jul/23
Answered by JDamian last updated on 23/Jul/23
Answered by Rasheed.Sindhi last updated on 23/Jul/23
$$\mathrm{log}_{\mathrm{3}} \mathrm{4}{x}+\mathrm{3log}_{\mathrm{27}} {x}=−\mathrm{6}\:\: \\ $$$$\mathrm{log}_{\mathrm{3}} \mathrm{4}{x}+\mathrm{3}\left(\frac{\mathrm{log}_{\mathrm{3}} \:{x}}{\mathrm{log}_{\mathrm{3}} \mathrm{27}\:}\right)=−\mathrm{6} \\ $$$$\mathrm{log}_{\mathrm{3}} \mathrm{4}{x}+\mathrm{3}\left(\frac{\mathrm{log}_{\mathrm{3}} \:{x}}{\mathrm{3}\:}\right)=−\mathrm{6} \\ $$$$\mathrm{log}_{\mathrm{3}} \mathrm{4}{x}+\mathrm{log}_{\mathrm{3}} \:{x}=−\mathrm{6} \\ $$$$\mathrm{log}_{\mathrm{3}} \left(\mathrm{4}{x}^{\mathrm{2}} \right)=−\mathrm{6} \\ $$$$\:\:\:\:\mathrm{3}^{−\mathrm{6}} =\mathrm{4}{x}^{\mathrm{2}} \: \\ $$$$\:\:\:{x}^{\mathrm{2}} =\frac{\mathrm{3}^{−\mathrm{6}} }{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}\centerdot\mathrm{3}^{\mathrm{6}} } \\ $$$$\:\:\:{x}=\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{54}} \\ $$