x-x-2-2-x-gt-0-

Question Number 84384 by M±th+et£s last updated on 12/Mar/20

{[x]}^{x} = 2 \sqrt{2}, \forall x > 0

Commented by mr W last updated on 12/Mar/20

[x] ⩽ x

{[x]}^{x} ⩽ x^{x}

f o r x^{x} = 2 \sqrt{2} w e g e t x = \frac{\ln 2 \sqrt{2}}{W (\ln 2 \sqrt{2})} \approx 1.6818

\Rightarrow f o r {[x]}^{x} = 2 \sqrt{2} w e h a v e \frac{\ln 2 \sqrt{2}}{W (\ln 2 \sqrt{2})} < x < 2

i . e . [x] = 1 a n d {[x]}^{x} = 1^{x} = 1 = 2 \sqrt{2}!

t h i s i s i m p o s s i b l e, i . e . {[x]}^{x} = 2 \sqrt{2} h a s n o

s o l u t i o n .

Commented by mr W last updated on 12/Mar/20

g e n e r a l l y,

f o r 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 . t h e r e i s s o l u t i o n f o r {[x]}^{x} = a i f

k^{k} ⩽ a < k^{k + 1}, b u t

i f k^{k + 1} ⩽ a < {(k + 1)}^{k + 1} t h e r e i s n o

s o l u t i o n f o r {[x]}^{x} = a .

k = 0 : 0 < a < 1

k = 1 : 1 < a < 4

k = 2 : 8 ⩽ a < 27

\dots \dots

2 \sqrt{2} \in (1, 4) \Rightarrow {[x]}^{x} = 2 \sqrt{2} h a s n o s o l u t i o n .

10 \in [8, 27) \Rightarrow {[x]}^{x} = 10 h a s n o s o l u t i o n .

Commented by mr W last updated on 12/Mar/20

e x a m p l e : s o l v e {[x]}^{x} = 3 \sqrt{2}

s i n c e 3 \sqrt{2} \in [4, 8) \Rightarrow t h e r e i s a s o l u t i o n!

a n d 2 < x < 3

\Rightarrow [x] = 2

\Rightarrow {[x]}^{x} = 2^{x} = 3 \sqrt{2}

\Rightarrow x = \frac{\ln 3 \sqrt{2}}{\ln 2} \approx 2.085

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