Question Number 162814 by amin96 last updated on 01/Jan/22
Answered by MathsFan last updated on 01/Jan/22
$$\:{let}\:\:{log}_{\mathrm{3}} {x}={u}\:\rightarrow\:{x}=\mathrm{3}^{{u}} \\ $$$$\:\mathrm{2}^{{u}} =\sqrt{\mathrm{3}^{{u}} }+\mathrm{1} \\ $$$$\:\mathrm{2}^{{u}} −\sqrt{\mathrm{3}^{{u}} }=\mathrm{4}−\mathrm{3} \\ $$$$\:\mathrm{2}^{{u}} =\mathrm{4}\:\:\rightarrow\:\:{u}=\mathrm{2} \\ $$$$\:{x}=\mathrm{3}^{\mathrm{2}} \: \\ $$$$\:{x}=\mathrm{9} \\ $$$$\: \\ $$
Answered by Exauce last updated on 01/Jan/22
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{X}}\:=\:^{\boldsymbol{{log}}_{\mathrm{3}} \mathrm{2}} \sqrt{\sqrt{\boldsymbol{{X}}}\:+\:\mathrm{1}} \\ $$