Question Number 210371 by Abdullahrussell last updated on 08/Aug/24
Answered by Spillover last updated on 08/Aug/24
$${use}\:{graphical}\:{method}\:{then}\:{find}\:{point}\:{of} \\ $$$${intersection} \\ $$$${let}\: \\ $$$${y}=^{{x}} \sqrt{\frac{\mathrm{2}}{{x}−\mathrm{1}}}\:\:\:\:\:\:\:\:\:\:\:\:{draw}\:{it}\:{graph} \\ $$$$\:{y}=\left({x}−\mathrm{2}\right)^{\left({x}−\mathrm{2}\right)} \:\:\:\:\:{draw}\:{it}\:{graph} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
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
$$\mathrm{Let}\:{x}>\mathrm{1} \\ $$$$\mathrm{2}^{\frac{\mathrm{1}}{{x}}} \left({x}−\mathrm{1}\right)^{−\frac{\mathrm{1}}{{x}}} =\left({x}−\mathrm{1}\right)^{{x}−\mathrm{2}} \\ $$$$\mathrm{2}^{\frac{\mathrm{1}}{{x}}} =\left({x}−\mathrm{1}\right)^{\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{{x}}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{{x}}=\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{ln}\:\left({x}−\mathrm{1}\right)}{{x}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\left({x}−\mathrm{1}\right)}=\left({x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}>\mathrm{1}\:\Rightarrow\:\mathrm{let}\:{x}=\mathrm{1}+\sqrt{{t}} \\ $$$$\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{ln}\:{t}}={t} \\ $$$$\mathrm{2n}\:\mathrm{2}\:={t}\mathrm{ln}\:{t} \\ $$$${t}=\mathrm{2} \\ $$$${x}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$
Commented by Spillover last updated on 08/Aug/24
$${perfect} \\ $$
Answered by mr W last updated on 08/Aug/24
$${x}>\mathrm{0} \\ $$$$\frac{\mathrm{2}}{{x}−\mathrm{1}}=\left({x}−\mathrm{1}\right)^{{x}\left({x}−\mathrm{2}\right)} \\ $$$$\mathrm{2}=\left({x}−\mathrm{1}\right)^{{x}\left({x}−\mathrm{2}\right)+\mathrm{1}} \\ $$$$\mathrm{2}=\left({x}−\mathrm{1}\right)^{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}^{\mathrm{2}} =\left[\left({x}−\mathrm{1}\right)^{\mathrm{2}} \right]^{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$
Commented by ajfour last updated on 08/Aug/24
Absolutely. Hello sir mrw
Commented by mr W last updated on 08/Aug/24
$${hi}!\:{nice}\:{to}\:{see}\:{you}\:{again}!\:{hope}\:{you} \\ $$$${are}\:{doing}\:{well}! \\ $$
Commented by ajfour last updated on 08/Aug/24
$${yes},\:{thanks}! \\ $$
Commented by Spillover last updated on 08/Aug/24
$${great} \\ $$