Question Number 84384 by M±th+et£s last updated on 12/Mar/20
![[x]^x =2(√2) , ∀x>0](https://www.tinkutara.com/question/Q84384.png)
$$\left[{x}\right]^{{x}} =\mathrm{2}\sqrt{\mathrm{2}}\:\:,\:\forall{x}>\mathrm{0} \\ $$
Commented by mr W last updated on 12/Mar/20
![[x]≤x [x]^x ≤x^x for x^x =2(√2) we get x=((ln 2(√2))/(W(ln 2(√2))))≈1.6818 ⇒for [x]^x =2(√2) we have ((ln 2(√2))/(W(ln 2(√2))))<x<2 i.e. [x]=1 and [x]^x =1^x =1=2(√2) ! this is impossible, i.e. [x]^x =2(√2) has no solution.](https://www.tinkutara.com/question/Q84389.png)
$$\left[{x}\right]\leqslant{x} \\ $$$$\left[{x}\right]^{{x}} \leqslant{x}^{{x}} \\ $$$${for}\:{x}^{{x}} =\mathrm{2}\sqrt{\mathrm{2}}\:{we}\:{get}\:{x}=\frac{\mathrm{ln}\:\mathrm{2}\sqrt{\mathrm{2}}}{{W}\left(\mathrm{ln}\:\mathrm{2}\sqrt{\mathrm{2}}\right)}\approx\mathrm{1}.\mathrm{6818} \\ $$$$\Rightarrow{for}\:\left[{x}\right]^{{x}} =\mathrm{2}\sqrt{\mathrm{2}}\:{we}\:{have}\:\:\frac{\mathrm{ln}\:\mathrm{2}\sqrt{\mathrm{2}}}{{W}\left(\mathrm{ln}\:\mathrm{2}\sqrt{\mathrm{2}}\right)}<{x}<\mathrm{2} \\ $$$${i}.{e}.\:\left[{x}\right]=\mathrm{1}\:{and}\:\left[{x}\right]^{{x}} =\mathrm{1}^{{x}} =\mathrm{1}=\mathrm{2}\sqrt{\mathrm{2}}\:! \\ $$$${this}\:{is}\:{impossible},\:{i}.{e}.\:\left[{x}\right]^{{x}} =\mathrm{2}\sqrt{\mathrm{2}}\:{has}\:{no} \\ $$$${solution}. \\ $$
Commented by mr W last updated on 12/Mar/20
![generally, for k≤x<k+1, i.e. x=k+d, 0≤d<1 [x]=k [x]^x =k^(k+d) ≥k^k [x]^x =k^(k+d) <k^(k+1) i.e. there is solution for [x]^x =a if k^k ≤a<k^(k+1) , but if k^(k+1) ≤a<(k+1)^(k+1) there is no solution for [x]^x =a. k=0: 0<a<1 k=1: 1<a<4 k=2: 8≤a<27 ...... 2(√2)∈(1,4) ⇒ [x]^x =2(√2) has no solution. 10∈[8,27) ⇒ [x]^x =10 has no solution.](https://www.tinkutara.com/question/Q84392.png)
$${generally}, \\ $$$${for}\:{k}\leqslant{x}<{k}+\mathrm{1},\:{i}.{e}.\:{x}={k}+{d},\:\mathrm{0}\leqslant{d}<\mathrm{1} \\ $$$$\left[{x}\right]={k} \\ $$$$\left[{x}\right]^{{x}} ={k}^{{k}+{d}} \geqslant{k}^{{k}} \\ $$$$\left[{x}\right]^{{x}} ={k}^{{k}+{d}} <{k}^{{k}+\mathrm{1}} \\ $$$${i}.{e}.\:{there}\:{is}\:{solution}\:{for}\:\left[{x}\right]^{{x}} ={a}\:{if} \\ $$$${k}^{{k}} \leqslant{a}<{k}^{{k}+\mathrm{1}} ,\:{but} \\ $$$${if}\:{k}^{{k}+\mathrm{1}} \leqslant{a}<\left({k}+\mathrm{1}\right)^{{k}+\mathrm{1}} \:{there}\:{is}\:{no} \\ $$$${solution}\:{for}\:\left[{x}\right]^{{x}} ={a}. \\ $$$${k}=\mathrm{0}:\:\mathrm{0}<{a}<\mathrm{1} \\ $$$${k}=\mathrm{1}:\:\mathrm{1}<{a}<\mathrm{4} \\ $$$${k}=\mathrm{2}:\:\mathrm{8}\leqslant{a}<\mathrm{27} \\ $$$$…… \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}\in\left(\mathrm{1},\mathrm{4}\right)\:\Rightarrow\:\left[{x}\right]^{{x}} =\mathrm{2}\sqrt{\mathrm{2}}\:{has}\:{no}\:{solution}. \\ $$$$\mathrm{10}\in\left[\mathrm{8},\mathrm{27}\right)\:\Rightarrow\:\left[{x}\right]^{{x}} =\mathrm{10}\:{has}\:{no}\:{solution}. \\ $$
Commented by mr W last updated on 12/Mar/20
![example: solve [x]^x =3(√2) since 3(√2)∈[4,8) ⇒there is a solution! and 2<x<3 ⇒[x]=2 ⇒[x]^x =2^x =3(√2) ⇒x=((ln 3(√2))/(ln 2))≈2.085](https://www.tinkutara.com/question/Q84402.png)
$${example}:\:{solve}\:\left[{x}\right]^{{x}} =\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${since}\:\mathrm{3}\sqrt{\mathrm{2}}\in\left[\mathrm{4},\mathrm{8}\right)\:\Rightarrow{there}\:{is}\:{a}\:{solution}! \\ $$$${and}\:\mathrm{2}<{x}<\mathrm{3} \\ $$$$\Rightarrow\left[{x}\right]=\mathrm{2} \\ $$$$\Rightarrow\left[{x}\right]^{{x}} =\mathrm{2}^{{x}} =\mathrm{3}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{ln}\:\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{ln}\:\mathrm{2}}\approx\mathrm{2}.\mathrm{085} \\ $$