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Question-209474




Question Number 209474 by Ismoiljon_008 last updated on 10/Jul/24
Commented by Ismoiljon_008 last updated on 10/Jul/24
help me please
$${help}\:{me}\:{please} \\ $$
Answered by Frix last updated on 10/Jul/24
t^t =2^(10(√2))   t=2^u   (2^u )^2^u  =2^(10(√2))   2^u ln 2^u  =10(√2)ln 2  2^u u=10(√2)=10×2^(1/2) =5×2^(3/2) =(5/2)×2^(5/2)   u=(5/2)  t=2^(5/2) =2^(2+(1/2)) =4(√2)
$${t}^{{t}} =\mathrm{2}^{\mathrm{10}\sqrt{\mathrm{2}}} \\ $$$${t}=\mathrm{2}^{{u}} \\ $$$$\left(\mathrm{2}^{{u}} \right)^{\mathrm{2}^{{u}} } =\mathrm{2}^{\mathrm{10}\sqrt{\mathrm{2}}} \\ $$$$\mathrm{2}^{{u}} \mathrm{ln}\:\mathrm{2}^{{u}} \:=\mathrm{10}\sqrt{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$$\mathrm{2}^{{u}} {u}=\mathrm{10}\sqrt{\mathrm{2}}=\mathrm{10}×\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{5}×\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\mathrm{5}}{\mathrm{2}}×\mathrm{2}^{\frac{\mathrm{5}}{\mathrm{2}}} \\ $$$${u}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${t}=\mathrm{2}^{\frac{\mathrm{5}}{\mathrm{2}}} =\mathrm{2}^{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{4}\sqrt{\mathrm{2}} \\ $$
Commented by Ismoiljon_008 last updated on 11/Jul/24
thanks
$${thanks} \\ $$
Answered by Kademi last updated on 11/Jul/24
        t^t  = 2^(10(√2))        tln t = 10(√2)ln 2       W(ln te^(ln t) ) = W((5/2)ln 2e^((5/2)ln 2)  )       ln t = ((5ln 2)/2)       t = e^((5ln 2)/2)        t = 4(√2)
$$\: \\ $$$$\:\:\:\:\:{t}^{{t}} \:=\:\mathrm{2}^{\mathrm{10}\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:{t}\mathrm{ln}\:{t}\:=\:\mathrm{10}\sqrt{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$$\:\:\:\:\:{W}\left(\mathrm{ln}\:{te}^{\mathrm{ln}\:{t}} \right)\:=\:{W}\left(\frac{\mathrm{5}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}{e}^{\frac{\mathrm{5}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}} \:\right) \\ $$$$\:\:\:\:\:\mathrm{ln}\:{t}\:=\:\frac{\mathrm{5ln}\:\mathrm{2}}{\mathrm{2}} \\ $$$$\:\:\:\:\:{t}\:=\:{e}^{\frac{\mathrm{5ln}\:\mathrm{2}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:{t}\:=\:\mathrm{4}\sqrt{\mathrm{2}} \\ $$

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