Menu Close

x-1-1-x-1-x-1-1-x-1-x-real-

Tinku Tara 0 Comments
Pin




Question Number 221544 by gregori last updated on 07/Jun/25
((x−1))^(1/(x+1)) = ((x+1))^(1/(x−1)) , x real
$$\:\:\sqrt[{{x}+\mathrm{1}}]{{x}−\mathrm{1}}\:=\:\sqrt[{{x}−\mathrm{1}}]{{x}+\mathrm{1}}\:,\:{x}\:{real}\: \\ $$
Answered by mr W last updated on 07/Jun/25
at first let′s have a look at the function f(x)=xln x. f′(x)=ln x+1=0 ⇒x=(1/e) for 0<x<(1/e) f(x) is decreasing, for (1/e)<x f(x) is increasing. the graph of f(x)=xln (x) see diagram below. (x−1)^(1/(x+1)) =(x+1)^(1/(x−1)) (x−1)ln (x−1)=(x+1)ln (x+1) ⇒f(x−1)=f(x+1) if this has a real root, then 0<x_1 =x−1<(1/e) ⇒1<x<1+(1/e) ...(i) (1/e)<x_2 =x+1<1 ⇒−1+(1/e)<x<0 ...(ii) (i) and (ii) are contridiction, that means f(x−1)=f(x+1) has no real root, i.e. ((x−1))^(1/(x+1)) =((x+1))^(1/(x−1)) has no real solution.
$${at}\:{first}\:{let}'{s}\:{have}\:{a}\:{look}\:{at}\:{the} \\ $$$${function}\:{f}\left({x}\right)={x}\mathrm{ln}\:{x}. \\ $$$${f}'\left({x}\right)=\mathrm{ln}\:{x}+\mathrm{1}=\mathrm{0}\:\Rightarrow{x}=\frac{\mathrm{1}}{{e}} \\ $$$${for}\:\mathrm{0}<{x}<\frac{\mathrm{1}}{{e}}\:{f}\left({x}\right)\:{is}\:{decreasing}, \\ $$$${for}\:\frac{\mathrm{1}}{{e}}<{x}\:\:{f}\left({x}\right)\:{is}\:{increasing}. \\ $$$${the}\:{graph}\:{of}\:{f}\left({x}\right)={x}\mathrm{ln}\:\left({x}\right)\:{see} \\ $$$${diagram}\:{below}. \\ $$$$\left({x}−\mathrm{1}\right)^{\frac{\mathrm{1}}{{x}+\mathrm{1}}} =\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{{x}−\mathrm{1}}} \\ $$$$\left({x}−\mathrm{1}\right)\mathrm{ln}\:\left({x}−\mathrm{1}\right)=\left({x}+\mathrm{1}\right)\mathrm{ln}\:\left({x}+\mathrm{1}\right) \\ $$$$\Rightarrow{f}\left({x}−\mathrm{1}\right)={f}\left({x}+\mathrm{1}\right) \\ $$$${if}\:{this}\:{has}\:{a}\:{real}\:{root},\:{then} \\ $$$$\mathrm{0}<{x}_{\mathrm{1}} ={x}−\mathrm{1}<\frac{\mathrm{1}}{{e}}\:\Rightarrow\mathrm{1}<{x}<\mathrm{1}+\frac{\mathrm{1}}{{e}}\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{1}}{{e}}<{x}_{\mathrm{2}} ={x}+\mathrm{1}<\mathrm{1}\:\Rightarrow−\mathrm{1}+\frac{\mathrm{1}}{{e}}<{x}<\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)\:{and}\:\left({ii}\right)\:{are}\:{contridiction},\:{that} \\ $$$${means}\:{f}\left({x}−\mathrm{1}\right)={f}\left({x}+\mathrm{1}\right)\:{has}\:{no} \\ $$$${real}\:{root},\:{i}.{e}. \\ $$$$\sqrt[{{x}+\mathrm{1}}]{{x}−\mathrm{1}}=\sqrt[{{x}−\mathrm{1}}]{{x}+\mathrm{1}}\:{has}\:{no}\:{real}\:{solution}. \\ $$
Commented by mr W last updated on 07/Jun/25
Commented by Frix last updated on 07/Jun/25
I think there′s no solution ∈C
$$\mathrm{I}\:\mathrm{think}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{solution}\:\in\mathbb{C} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *