What is the answer to x^x=sqrt(2) ?

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

Solve for x: x^x=sqrt(2)

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

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x^{x} = 2

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

Decimal Notation

x = 1.30435 \dots

Solution steps

x^{x} = 2

Prepare

x^{x} = 2

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Rewrite

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

in Lambert form:

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

Solve

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

Verify Solutions:

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

True

The solution is

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

Substitute back

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

solve for

x

Solve

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

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

1 6^{x - 4} = 25 6^{3 x + 3}

7^{2 x - 1} = 49 \cdot 7^{x - 1}

3^{2 x} - 5 (3)^{x} + 4 = 0

3^{5 x - 7} = 21

5^{x} + 5^{x - 1} + 5^{x - 2} = 31

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