Question-210371

Question Number 210371 by Abdullahrussell last updated on 08/Aug/24

Answered by Spillover last updated on 08/Aug/24

use graphical method then find point of intersection let y=^x (√(2/(x−1))) draw it graph y=(x−2)^((x−2)) draw it graph

$${use}\:{graphical}\:{method}\:{then}\:{find}\:{point}\:{of} \\ $$$${intersection} \\ $$$${let}\: \\ $$$${y}=^{{x}} \sqrt{\frac{\mathrm{2}}{{x}−\mathrm{1}}}\:\:\:\:\:\:\:\:\:\:\:\:{draw}\:{it}\:{graph} \\ $$$$\:{y}=\left({x}−\mathrm{2}\right)^{\left({x}−\mathrm{2}\right)} \:\:\:\:\:{draw}\:{it}\:{graph} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by Spillover last updated on 08/Aug/24

Commented by Spillover last updated on 08/Aug/24

$$ \\ $$The point where the two curves intersect represents the solution to the equation. From the graph, you can see that the intersection occurs at approximately x=2.414

Answered by Frix last updated on 08/Aug/24

Let x>1 2^(1/x) (x−1)^(−(1/x)) =(x−1)^(x−2) 2^(1/x) =(x−1)^(((x−1)^2 )/x) ((ln 2)/x)=(((x−1)^2 ln (x−1))/x) ((ln 2)/(ln (x−1)))=(x−1)^2 x>1 ⇒ let x=1+(√t) ((2ln 2)/(ln t))=t 2n 2 =tln t t=2 x=1+(√2)

$$\mathrm{Let}\:{x}>\mathrm{1} \\ $$$$\mathrm{2}^{\frac{\mathrm{1}}{{x}}} \left({x}−\mathrm{1}\right)^{−\frac{\mathrm{1}}{{x}}} =\left({x}−\mathrm{1}\right)^{{x}−\mathrm{2}} \\ $$$$\mathrm{2}^{\frac{\mathrm{1}}{{x}}} =\left({x}−\mathrm{1}\right)^{\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{{x}}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{{x}}=\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{ln}\:\left({x}−\mathrm{1}\right)}{{x}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\left({x}−\mathrm{1}\right)}=\left({x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}>\mathrm{1}\:\Rightarrow\:\mathrm{let}\:{x}=\mathrm{1}+\sqrt{{t}} \\ $$$$\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{ln}\:{t}}={t} \\ $$$$\mathrm{2n}\:\mathrm{2}\:={t}\mathrm{ln}\:{t} \\ $$$${t}=\mathrm{2} \\ $$$${x}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$

Commented by Spillover last updated on 08/Aug/24

$${perfect} \\ $$

Answered by mr W last updated on 08/Aug/24

x>0 (2/(x−1))=(x−1)^(x(x−2)) 2=(x−1)^(x(x−2)+1) 2=(x−1)^((x−1)^2 ) 2^2 =[(x−1)^2 ]^((x−1)^2 ) ⇒(x−1)^2 =2 ⇒x=1+(√2)

$${x}>\mathrm{0} \\ $$$$\frac{\mathrm{2}}{{x}−\mathrm{1}}=\left({x}−\mathrm{1}\right)^{{x}\left({x}−\mathrm{2}\right)} \\ $$$$\mathrm{2}=\left({x}−\mathrm{1}\right)^{{x}\left({x}−\mathrm{2}\right)+\mathrm{1}} \\ $$$$\mathrm{2}=\left({x}−\mathrm{1}\right)^{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}^{\mathrm{2}} =\left[\left({x}−\mathrm{1}\right)^{\mathrm{2}} \right]^{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$

Commented by ajfour last updated on 08/Aug/24

Absolutely. Hello sir mrw

Commented by mr W last updated on 08/Aug/24