What is the answer to x^x=7 ?

The answer to x^x=7 is x=(ln(7))/(\W_{0)(ln(7))}

The solution is x=(ln(7))/(\W_{0)(ln(7))}

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x^x=7

x^{x} = 7

x = \frac{ln ( 7 )}{W _{0} ( ln ( 7 ) )}

Solution steps

x^{x} = 7

Prepare

x^{x} = 7

for Lambert form:

x e^{- \frac{l n ( 7 )}{x}} = 1

Rewrite the equation with

\frac{ln ( 7 )}{x} = u

and

x = \frac{ln ( 7 )}{u}

(\frac{ln ( 7 )}{u}) e^{- u} = 1

Rewrite

(\frac{ln ( 7 )}{u}) e^{- u} = 1

in Lambert form:

u e^{u} = ln (7)

Solve

u e^{u} = ln (7) : u = W_{0} (ln (7))

Verify Solutions:

u = W_{0} (ln (7))

True

The solution is

u = W_{0} (ln (7))

Substitute back

u = \frac{ln ( 7 )}{x},

solve for

x

Solve

\frac{ln ( 7 )}{x} = W_{0} (ln (7)) : x = \frac{ln ( 7 )}{W _{0} ( ln ( 7 ) )}

x = \frac{ln ( 7 )}{W _{0} ( ln ( 7 ) )}

2 x = e^{- x}

e^{- l n (x)} = 20

2^{6} \cdot 2^{12} = (2^{x})^{2}

2^{x} = \frac{32}{8}

e^{x} + 24 e^{- x} = 10