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Question Number 193793 by Mingma last updated on 20/Jun/23

Commented by MM42 last updated on 20/Jun/23

$$yes.youareright$$$$itwasmymistake.thankyou$$

Commented by Frix last updated on 20/Jun/23

$$1.61053599<x<1.63195612$$

Commented by Frix last updated on 20/Jun/23

$$\lfloor 3.14\rfloor =3\mathrm{but}\lceil 3.14\rceil =4$$$$$$$$x=1.62$$$$\lfloor 1.{62}^{\lceil 1.{62}^{\lceil 1.62\rceil }\rceil 1.62}\rfloor =\lfloor 1.{62}^{\lceil 1.{62}^2\rceil 1.62}\rfloor =$$$$=\lfloor 1.{62}^{3\times 1.62}\rfloor =\lfloor 10.429...\rfloor =10$$

Commented by Mingma last updated on 20/Jun/23

perfect ��

Answered by mr W last updated on 22/Jun/23

$$ifx⩾2:$$$$\lfloor x^{\lceil x^{\lceil x\rceil }\rceil x}\rfloor ⩾2^{2^2\times 2}=2^8=256>10$$$$ifx=1:$$$$\lfloor x^{\lceil x^{\lceil x\rceil }\rceil x}\rfloor =1<10$$$$\Rightarrow 1<x<2,i.e.\lceil x\rceil =2$$$$ifx⩽\sqrt{2}:$$$$\lfloor x^{\lceil x^{\lceil x\rceil }\rceil x}\rfloor ⩽\lfloor {\left(\sqrt{2}\right)}^{2\sqrt{2}}\rfloor =2<10$$$$\Rightarrow x>\sqrt{2}$$$$\lceil x^{\lceil x\rceil }\rceil =3,i.e.2<x^2⩽3\Rightarrow x⩽\sqrt{3}$$$$\lfloor x^{\lceil x^{\lceil x\rceil }\rceil x}\rfloor =\lfloor x^{3x}\rfloor =10$$$$10⩽x^{3x}<11$$$${10}^{\frac{1}{3}}⩽x^x<{11}^{\frac{1}{3}}$$$$\Rightarrow \frac{\ln 10}{3W\left(\frac{\ln 10}{3}\right)}⩽x<\frac{\ln 11}{3W\left(\frac{\ln 11}{3}\right)}✓$$$$$$$$i.e.\approx 1.61053599⩽x<\approx 1.63195611$$

Commented by Frix last updated on 26/Jun/23

$$\mathrm{Great}!$$

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