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Question Number 139805 by mr W last updated on 01/May/21
solve for x∈R x^x^n =a with n, a∈R^+ find also the range of a such that a solution exists.
$${solve}\:{for}\:{x}\in\mathbb{R} \\ $$$$\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{n}}} } =\boldsymbol{{a}} \\ $$$${with}\:{n},\:{a}\in\mathbb{R}^{+} \\ $$$${find}\:{also}\:{the}\:{range}\:{of}\:\boldsymbol{{a}}\:{such}\:{that}\:{a} \\ $$$${solution}\:{exists}. \\ $$
Commented by mr W last updated on 01/May/21
i thought about this general equation for a long time and tried to find a formula to express the solution. we know if n=a, the solution for x^x^a =a is the same as for x^a =a, i.e. x=(a)^(1/a) . but what is the solution for the general case with n≠a?
$${i}\:{thought}\:{about}\:{this}\:{general}\:{equation} \\ $$$${for}\:{a}\:{long}\:{time}\:{and}\:{tried}\:{to}\:{find} \\ $$$${a}\:{formula}\:{to}\:{express}\:{the}\:{solution}. \\ $$$${we}\:{know}\:{if}\:{n}={a},\:{the}\:{solution}\:{for} \\ $$$${x}^{{x}^{{a}} } ={a}\:{is}\:{the}\:{same}\:{as}\:{for}\:{x}^{{a}} ={a},\:{i}.{e}. \\ $$$${x}=\sqrt[{{a}}]{{a}}.\:{but}\:{what}\:{is}\:{the}\:{solution}\:{for} \\ $$$${the}\:{general}\:{case}\:{with}\:{n}\neq{a}? \\ $$
Answered by mr W last updated on 02/May/21
here the solution i got for the general case x^x^n =a. x^x^n =a x=a^(1/x^n ) =e^((ln a)/x^n ) x^n =e^((nln a)/x^n ) (1/x^n )e^((nln a)/x^n ) =1 ((nln a)/x^n )e^((nln a)/x^n ) =nln a ((nln a)/x^n )=W(nln a) x^n =((nln a)/(W(nln a))) ⇒x=(((nln a)/(W(nln a))))^(1/n) such that a real root exists, W(nln a) must be defined. this is the case if nln a≥−(1/e), i.e. a≥e^(−(1/(ne))) =(1/( (e)^(1/(ne)) )). therefore: for a<(1/( (e)^(1/(ne)) )) ⇒ no root for a=(1/( (e)^(1/(ne)) )) ⇒ one root for (1/( (e)^(1/(ne)) ))<a<1 ⇒ two roots for a≥1 ⇒ one root example: x^x^5 =10 ⇒x=(((5ln 10)/(W(5ln 10))))^(1/5) ≈(((5ln 10)/(1.835928)))^(1/5) =1.443664 example: x^x^3 =(9/(10)) ⇒x=(((3ln 0.9)/(W(3ln 0.9))))^(1/3) ≈ { (((((3ln 0.9)/(−1.656404)))^(1/3) =0.575719)),(((((3ln 0.9)/(−0.545257)))^(1/3) =0.833808)) :}
$${here}\:{the}\:{solution}\:{i}\:{got}\:{for}\:{the} \\ $$$${general}\:{case}\:{x}^{{x}^{{n}} } ={a}. \\ $$$${x}^{{x}^{{n}} } ={a} \\ $$$${x}={a}^{\frac{\mathrm{1}}{{x}^{{n}} }} ={e}^{\frac{\mathrm{ln}\:{a}}{{x}^{{n}} }} \\ $$$${x}^{{n}} ={e}^{\frac{{n}\mathrm{ln}\:{a}}{{x}^{{n}} }} \\ $$$$\frac{\mathrm{1}}{{x}^{{n}} }{e}^{\frac{{n}\mathrm{ln}\:{a}}{{x}^{{n}} }} =\mathrm{1} \\ $$$$\frac{{n}\mathrm{ln}\:{a}}{{x}^{{n}} }{e}^{\frac{{n}\mathrm{ln}\:{a}}{{x}^{{n}} }} ={n}\mathrm{ln}\:{a} \\ $$$$\frac{{n}\mathrm{ln}\:{a}}{{x}^{{n}} }=\mathbb{W}\left({n}\mathrm{ln}\:{a}\right) \\ $$$${x}^{{n}} =\frac{{n}\mathrm{ln}\:{a}}{\mathbb{W}\left({n}\mathrm{ln}\:{a}\right)} \\ $$$$\Rightarrow{x}=\sqrt[{{n}}]{\frac{{n}\mathrm{ln}\:{a}}{\mathbb{W}\left({n}\mathrm{ln}\:{a}\right)}} \\ $$$${such}\:{that}\:{a}\:{real}\:{root}\:{exists},\:\mathbb{W}\left({n}\mathrm{ln}\:{a}\right) \\ $$$${must}\:{be}\:{defined}.\:{this}\:{is}\:{the}\:{case}\:{if} \\ $$$${n}\mathrm{ln}\:{a}\geqslant−\frac{\mathrm{1}}{{e}},\:{i}.{e}.\:{a}\geqslant{e}^{−\frac{\mathrm{1}}{{ne}}} =\frac{\mathrm{1}}{\:\sqrt[{{ne}}]{{e}}}. \\ $$$${therefore}: \\ $$$${for}\:{a}<\frac{\mathrm{1}}{\:\sqrt[{{ne}}]{{e}}}\:\Rightarrow\:{no}\:{root} \\ $$$${for}\:{a}=\frac{\mathrm{1}}{\:\sqrt[{{ne}}]{{e}}}\:\Rightarrow\:{one}\:{root} \\ $$$${for}\:\frac{\mathrm{1}}{\:\sqrt[{{ne}}]{{e}}}<{a}<\mathrm{1}\:\Rightarrow\:{two}\:{roots} \\ $$$${for}\:{a}\geqslant\mathrm{1}\:\Rightarrow\:{one}\:{root} \\ $$$$ \\ $$$${example}:\:{x}^{{x}^{\mathrm{5}} } =\mathrm{10} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{5}}]{\frac{\mathrm{5ln}\:\mathrm{10}}{\mathbb{W}\left(\mathrm{5ln}\:\mathrm{10}\right)}}\approx\sqrt[{\mathrm{5}}]{\frac{\mathrm{5ln}\:\mathrm{10}}{\mathrm{1}.\mathrm{835928}}}=\mathrm{1}.\mathrm{443664} \\ $$$$ \\ $$$${example}:\:{x}^{{x}^{\mathrm{3}} } =\frac{\mathrm{9}}{\mathrm{10}} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3ln}\:\mathrm{0}.\mathrm{9}}{\mathbb{W}\left(\mathrm{3ln}\:\mathrm{0}.\mathrm{9}\right)}}\approx\begin{cases}{\sqrt[{\mathrm{3}}]{\frac{\mathrm{3ln}\:\mathrm{0}.\mathrm{9}}{−\mathrm{1}.\mathrm{656404}}}=\mathrm{0}.\mathrm{575719}}\\{\sqrt[{\mathrm{3}}]{\frac{\mathrm{3ln}\:\mathrm{0}.\mathrm{9}}{−\mathrm{0}.\mathrm{545257}}}=\mathrm{0}.\mathrm{833808}}\end{cases} \\ $$
Commented by Tawa11 last updated on 23/Jul/21
great
$$\mathrm{great} \\ $$

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