What is the answer to x^x=2^{2084} ?

The answer to x^x=2^{2084} is x=(2084ln(2))/(\W_{0)(2084ln(2))}

Solve for x: x^x=2^{2084}

The solution is x=(2084ln(2))/(\W_{0)(2084ln(2))}

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x^{x} = 2^{2084}

x = \frac{2084 ln ( 2 )}{W _{0} ( 2084 ln ( 2 ) )}

Decimal Notation

x = 259.80816 \dots

Solution steps

x^{x} = 2^{2084}

Prepare

x^{x} = 2^{2084}

for Lambert form:

x e^{- \frac{2084 l n ( 2 )}{x}} = 1

Rewrite the equation with

\frac{2084 ln ( 2 )}{x} = u

and

x = \frac{2084 ln ( 2 )}{u}

(\frac{2084 ln ( 2 )}{u}) e^{- u} = 1

Rewrite

(\frac{2084 ln ( 2 )}{u}) e^{- u} = 1

in Lambert form:

u e^{u} = 2084 ln (2)

Solve

u e^{u} = 2084 ln (2) : u = W_{0} (2084 ln (2))

Verify Solutions:

u = W_{0} (2084 ln (2))

True

The solution is

u = W_{0} (2084 ln (2))

Substitute back

u = \frac{2084 ln ( 2 )}{x},

solve for

x

Solve

\frac{2084 ln ( 2 )}{x} = W_{0} (2084 ln (2)) : x = \frac{2084 ln ( 2 )}{W _{0} ( 2084 ln ( 2 ) )}

x = \frac{2084 ln ( 2 )}{W _{0} ( 2084 ln ( 2 ) )}

5^{x} - 24 = \frac{5 ^{2}}{5} x

5^{(x - 1)} + 5^{(x - 2)} = 30

(\frac{1}{2})^{\frac{t}{2}} = 64

3^{2 x - 1} \cdot 2 7^{3 - x} = 1

3 x + 2 = e^{x}

What is the answer to x^x=2^{2084} ?
The answer to x^x=2^{2084} is x=(2084ln(2))/(\W_{0)(2084ln(2))}
Solve for x: x^x=2^{2084}
The solution is x=(2084ln(2))/(\W_{0)(2084ln(2))}