Question Number 204142 by Tawa11 last updated on 06/Feb/24
$$\mathrm{Solve}:\:\:\mathrm{x}^{\mathrm{x}} \:\:=\:\:\mathrm{27}^{\mathrm{x}\:\:−\:\:\mathrm{3}} \\ $$
Answered by Frix last updated on 07/Feb/24
![x^x =27^(x−3) xln x =(x−3)ln 27 x(ln x −ln 27)=−3ln 3^3 xln (x/(27)) =−9ln 3 [obviously x=9 because 9ln (9/(27)) =9ln (1/3) =−9ln 3] x=27t 27tln t =−9ln 3 tln t =−((ln 3)/3) t=e^u e^u u=−((ln 3)/3) u=W(−((ln 3)/3)) t=e^(W(−((ln 3)/3))) x=27e^(W(−((ln 3)/3))) x=9∨x≈10.8956521444325](https://www.tinkutara.com/question/Q204144.png)
$${x}^{{x}} =\mathrm{27}^{{x}−\mathrm{3}} \\ $$$${x}\mathrm{ln}\:{x}\:=\left({x}−\mathrm{3}\right)\mathrm{ln}\:\mathrm{27} \\ $$$${x}\left(\mathrm{ln}\:{x}\:−\mathrm{ln}\:\mathrm{27}\right)=−\mathrm{3ln}\:\mathrm{3}^{\mathrm{3}} \\ $$$${x}\mathrm{ln}\:\frac{{x}}{\mathrm{27}}\:=−\mathrm{9ln}\:\mathrm{3} \\ $$$$\:\:\:\:\:\left[\mathrm{obviously}\:{x}=\mathrm{9}\:\mathrm{because}\:\mathrm{9ln}\:\frac{\mathrm{9}}{\mathrm{27}}\:=\mathrm{9ln}\:\frac{\mathrm{1}}{\mathrm{3}}\:=−\mathrm{9ln}\:\mathrm{3}\right] \\ $$$${x}=\mathrm{27}{t} \\ $$$$\mathrm{27}{t}\mathrm{ln}\:{t}\:=−\mathrm{9ln}\:\mathrm{3} \\ $$$${t}\mathrm{ln}\:{t}\:=−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}} \\ $$$${t}=\mathrm{e}^{{u}} \\ $$$$\mathrm{e}^{{u}} {u}=−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}} \\ $$$${u}=\mathrm{W}\left(−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}}\right) \\ $$$${t}=\mathrm{e}^{\mathrm{W}\left(−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}}\right)} \\ $$$${x}=\mathrm{27e}^{\mathrm{W}\left(−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}}\right)} \\ $$$${x}=\mathrm{9}\vee{x}\approx\mathrm{10}.\mathrm{8956521444325} \\ $$
Commented by Tawa11 last updated on 07/Feb/24
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$