Question Number 55806 by Tawa1 last updated on 04/Mar/19
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\mathrm{x}^{\mathrm{log}_{\mathrm{3}} \mathrm{2}} \:\:=\:\:\sqrt{\mathrm{x}}\:\:+\:\:\mathrm{1} \\ $$
Answered by MJS last updated on 04/Mar/19
![ln x^(log_3 2) =ln x^((ln 2)/(ln 3)) =((ln 2)/(ln 3))ln x = [let x=3^t ] =tln 2 e^(tln 2) =2^t 2^t =3^(t/2) +1 ⇒ for t=2n both sides are real 4^n =3^n +1 ⇒ n=1 ⇒ t=2 ⇒ x=9](Q55809.png)
$$\mathrm{ln}\:{x}^{\mathrm{log}_{\mathrm{3}} \:\mathrm{2}} =\mathrm{ln}\:{x}^{\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}} =\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}\mathrm{ln}\:{x}\:= \\ $$$$\:\:\:\:\:\left[\mathrm{let}\:{x}=\mathrm{3}^{{t}} \right] \\ $$$$={t}\mathrm{ln}\:\mathrm{2} \\ $$$$\mathrm{e}^{{t}\mathrm{ln}\:\mathrm{2}} =\mathrm{2}^{{t}} \\ $$$$\mathrm{2}^{{t}} =\mathrm{3}^{\frac{{t}}{\mathrm{2}}} +\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{for}\:{t}=\mathrm{2}{n}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{real} \\ $$$$\mathrm{4}^{{n}} =\mathrm{3}^{{n}} +\mathrm{1}\:\Rightarrow\:{n}=\mathrm{1}\:\Rightarrow\:{t}=\mathrm{2}\:\Rightarrow\:{x}=\mathrm{9} \\ $$
Commented by Tawa1 last updated on 04/Mar/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$