What is the answer to x^x=5^{2/25} ?

The answer to x^x=5^{2/25} is x=(2ln(5))/(25\W_{0)((2ln(5))/(25))}

Solve for x: x^x=5^{2/25}

The solution is x=(2ln(5))/(25\W_{0)((2ln(5))/(25))}

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x^{x} = 5^{\frac{2}{25}}

x = \frac{2 ln ( 5 )}{25 W _{0} ( \frac{2 l n ( 5 )}{25} )}

Decimal Notation

x = 1.12163 \dots

Solution steps

x^{x} = 5^{\frac{2}{25}}

Prepare

x^{x} = 5^{\frac{2}{25}}

for Lambert form:

x e^{- \frac{2 l n ( 5 )}{25 x}} = 1

Rewrite the equation with

\frac{2 ln ( 5 )}{25 x} = u

and

x = \frac{2 ln ( 5 )}{25 u}

(\frac{2 ln ( 5 )}{25 u}) e^{- u} = 1

Rewrite

(\frac{2 ln ( 5 )}{25 u}) e^{- u} = 1

in Lambert form:

e^{u} u = \frac{2 ln ( 5 )}{25}

Solve

e^{u} u = \frac{2 ln ( 5 )}{25} : u = W_{0} (\frac{2 ln ( 5 )}{25})

Verify Solutions:

u = W_{0} (\frac{2 ln ( 5 )}{25})

True

The solution is

u = W_{0} (\frac{2 ln ( 5 )}{25})

Substitute back

u = \frac{2 ln ( 5 )}{25 x},

solve for

x

Solve

\frac{2 ln ( 5 )}{25 x} = W_{0} (\frac{2 ln ( 5 )}{25}) : x = \frac{2 ln ( 5 )}{25 W _{0} ( \frac{2 l n ( 5 )}{25} )}

x = \frac{2 ln ( 5 )}{25 W _{0} ( \frac{2 l n ( 5 )}{25} )}

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

2 (10)^{x} + 1 0^{x + 1} = 4

67 = 2 1^{x}

y = b + a e^{- x}, x

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

What is the answer to x^x=5^{2/25} ?
The answer to x^x=5^{2/25} is x=(2ln(5))/(25\W_{0)((2ln(5))/(25))}
Solve for x: x^x=5^{2/25}
The solution is x=(2ln(5))/(25\W_{0)((2ln(5))/(25))}