What is the answer to x^x=\sqrt[9]{1/3} ?

The answer to x^x=\sqrt[9]{1/3} is x= 1/27 ,x=-(ln(3)}{9\W_{0}(-\frac{ln(3))/9)}

Solve for x: x^x=\sqrt[9]{1/3}

The solution is x= 1/27 ,x=-(ln(3)}{9\W_{0}(-\frac{ln(3))/9)}

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x = \frac{1}{27}, x = - \frac{ln ( 3 )}{9 W _{0} ( - \frac{l n ( 3 )}{9} )}

Decimal Notation

x = 0.03703 \dots, x = 0.86894 \dots

Solution steps

Prepare

for Lambert form:

x e^{\frac{l n ( 3 )}{9 x}} = 1

Rewrite the equation with

- \frac{ln ( 3 )}{9 x} = u

and

x = - \frac{ln ( 3 )}{9 u}

(- \frac{ln ( 3 )}{9 u}) e^{- u} = 1

Solve

(- \frac{ln ( 3 )}{9 u}) e^{- u} = 1 : u = - 3 ln (3), u = W_{0} (- \frac{ln ( 3 )}{9})

u = - 3 ln (3), u = W_{0} (- \frac{ln ( 3 )}{9})

Substitute back

u = - \frac{ln ( 3 )}{9 x},

solve for

x

Solve

- \frac{ln ( 3 )}{9 x} = - 3 ln (3) : x = \frac{1}{27}

Solve

- \frac{ln ( 3 )}{9 x} = W_{0} (- \frac{ln ( 3 )}{9}) : x = - \frac{ln ( 3 )}{9 W _{0} ( - \frac{l n ( 3 )}{9} )}

x = \frac{1}{27}, x = - \frac{ln ( 3 )}{9 W _{0} ( - \frac{l n ( 3 )}{9} )}

4^{x - 3} = 8^{4 - x}

2^{x + 1} = 128

2 5^{x - 5} = 625

4^{1 - 2 x} = 2

3^{4 x} = 9^{x + 1}

What is the answer to x^x=\sqrt[9]{1/3} ?
The answer to x^x=\sqrt[9]{1/3} is x= 1/27 ,x=-(ln(3)}{9\W_{0}(-\frac{ln(3))/9)}
Solve for x: x^x=\sqrt[9]{1/3}
The solution is x= 1/27 ,x=-(ln(3)}{9\W_{0}(-\frac{ln(3))/9)}