What is the answer to n^n=2^{1000000} ?

The answer to n^n=2^{1000000} is n=(1000000ln(2))/(\W_{0)(1000000ln(2))}

Solve for n: n^n=2^{1000000}

The solution is n=(1000000ln(2))/(\W_{0)(1000000ln(2))}

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n^{n} = 2^{1000000}

n = \frac{1000000 ln ( 2 )}{W _{0} ( 1000000 ln ( 2 ) )}

Decimal Notation

n = 62746.12646 \dots

Solution steps

n^{n} = 2^{1000000}

Prepare

n^{n} = 2^{1000000}

for Lambert form:

n e^{- \frac{1000000 l n ( 2 )}{n}} = 1

Rewrite the equation with

\frac{1000000 ln ( 2 )}{n} = u

and

n = \frac{1000000 ln ( 2 )}{u}

(\frac{1000000 ln ( 2 )}{u}) e^{- u} = 1

Rewrite

(\frac{1000000 ln ( 2 )}{u}) e^{- u} = 1

in Lambert form:

u e^{u} = 1000000 ln (2)

Solve

u e^{u} = 1000000 ln (2) : u = W_{0} (1000000 ln (2))

Verify Solutions:

u = W_{0} (1000000 ln (2))

True

The solution is

u = W_{0} (1000000 ln (2))

Substitute back

u = \frac{1000000 ln ( 2 )}{n},

solve for

n

Solve

\frac{1000000 ln ( 2 )}{n} = W_{0} (1000000 ln (2)) : n = \frac{1000000 ln ( 2 )}{W _{0} ( 1000000 ln ( 2 ) )}

n = \frac{1000000 ln ( 2 )}{W _{0} ( 1000000 ln ( 2 ) )}

3 \cdot 9^{2 x + 3} = 1

5^{x - 1} - 5^{x} = - 100

e^{(2 x)} - e^{(x)} - 12 = 0

\frac{2 ^{x + 1}}{4 ^{x - 1}} = 2 x

e^{2 x} - x^{3} = 0

What is the answer to n^n=2^{1000000} ?
The answer to n^n=2^{1000000} is n=(1000000ln(2))/(\W_{0)(1000000ln(2))}
Solve for n: n^n=2^{1000000}
The solution is n=(1000000ln(2))/(\W_{0)(1000000ln(2))}