What is the answer to x^x=sqrt((\sqrt{2))/2} ?

The answer to x^x=sqrt((\sqrt{2))/2} is x= 1/16 ,x=-(ln(2)}{4\W_{0}(-\frac{ln(2))/4)}

Solve for x: x^x=sqrt((\sqrt{2))/2}

The solution is x= 1/16 ,x=-(ln(2)}{4\W_{0}(-\frac{ln(2))/4)}

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x^{x} = \frac{2}{2}

x = \frac{1}{16}, x = - \frac{ln ( 2 )}{4 W _{0} ( - \frac{l n ( 2 )}{4} )}

Decimal Notation

x = 0.0625, x = 0.80669 \dots

Solution steps

x^{x} = \frac{2}{2}

Prepare

x^{x} = \frac{2}{2}

for Lambert form:

x e^{\frac{l n ( 2 )}{4 x}} = 1

Rewrite the equation with

- \frac{ln ( 2 )}{4 x} = u

and

x = - \frac{ln ( 2 )}{4 u}

(- \frac{ln ( 2 )}{4 u}) e^{- u} = 1

Solve

(- \frac{ln ( 2 )}{4 u}) e^{- u} = 1 : u = - 4 ln (2), u = W_{0} (- \frac{ln ( 2 )}{4})

u = - 4 ln (2), u = W_{0} (- \frac{ln ( 2 )}{4})

Substitute back

u = - \frac{ln ( 2 )}{4 x},

solve for

x

Solve

- \frac{ln ( 2 )}{4 x} = - 4 ln (2) : x = \frac{1}{16}

Solve

- \frac{ln ( 2 )}{4 x} = W_{0} (- \frac{ln ( 2 )}{4}) : x = - \frac{ln ( 2 )}{4 W _{0} ( - \frac{l n ( 2 )}{4} )}

x = \frac{1}{16}, x = - \frac{ln ( 2 )}{4 W _{0} ( - \frac{l n ( 2 )}{4} )}

25 6^{3 x + 3} = 102 4^{2 x + 2}

216 = 6^{2 r - 11}

2^{x - x^{2}} = \frac{1}{4 ^{x}}

62 5^{x} = \frac{1}{5}

e^{x^{2}} = 5 e^{2 x - 1}

What is the answer to x^x=sqrt((\sqrt{2))/2} ?
The answer to x^x=sqrt((\sqrt{2))/2} is x= 1/16 ,x=-(ln(2)}{4\W_{0}(-\frac{ln(2))/4)}
Solve for x: x^x=sqrt((\sqrt{2))/2}
The solution is x= 1/16 ,x=-(ln(2)}{4\W_{0}(-\frac{ln(2))/4)}