What is the answer to x^x=\sqrt[4]{2} ?

The answer to x^x=\sqrt[4]{2} is x=(ln(2)}{4\W_{0}(\frac{ln(2))/4)}

Solve for x: x^x=\sqrt[4]{2}

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

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

Solution steps

Prepare

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

Rewrite

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

in Lambert form:

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

Solve

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

Verify Solutions:

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

True

The solution is

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

Substitute back

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

solve for

x

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{ln ( 2 )}{4 W _{0} ( \frac{l n ( 2 )}{4} )}

- 9 \cdot 2 7^{x} = 27

5^{3 x - 1} = (\frac{1}{25})^{2 x + 3}

(\frac{1}{3})^{x} = 52

\frac{9 ^{x^{2} - x}}{3 ^{3 x - 2}} = 1

4^{(1 - x)} = 3^{(2 x + 3)}

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