What is the answer to n=2^{(n/8)} ?

The answer to n=2^{(n/8)} is n=-(8\W_{-1}(-(ln(2))/8))/(ln(2)),n=-(8\W_{0}(-(ln(2))/8))/(ln(2))

Solve for n: n=2^{(n/8)}

The solution is n=-(8\W_{-1}(-(ln(2))/8))/(ln(2)),n=-(8\W_{0}(-(ln(2))/8))/(ln(2))

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n = 2^{(\frac{n}{8})}

n = - \frac{8 W _{- 1} ( - \frac{l n ( 2 )}{8} )}{ln ( 2 )}, n = - \frac{8 W _{0} ( - \frac{l n ( 2 )}{8} )}{ln ( 2 )}

Decimal Notation

n = 43.55926 \dots, n = 1.09999 \dots

Solution steps

n = 2^{(\frac{n}{8})}

Prepare

n = 2^{(\frac{n}{8})}

for Lambert form:

n e^{- \frac{l n ( 2 ) n}{8}} = 1

Rewrite the equation with

- \frac{ln ( 2 ) n}{8} = u

and

n = - \frac{8 u}{ln ( 2 )}

(- \frac{8 u}{ln ( 2 )}) e^{u} = 1

Rewrite

(- \frac{8 u}{ln ( 2 )}) e^{u} = 1

in Lambert form:

e^{u} u = - \frac{ln ( 2 )}{8}

Solve

e^{u} u = - \frac{ln ( 2 )}{8} : u = W_{- 1} (- \frac{ln ( 2 )}{8}), u = W_{0} (- \frac{ln ( 2 )}{8})

u = W_{- 1} (- \frac{ln ( 2 )}{8}), u = W_{0} (- \frac{ln ( 2 )}{8})

Substitute back

u = - \frac{ln ( 2 ) n}{8},

solve for

n

Solve

- \frac{ln ( 2 ) n}{8} = W_{- 1} (- \frac{ln ( 2 )}{8}) : n = - \frac{8 W _{- 1} ( - \frac{l n ( 2 )}{8} )}{ln ( 2 )}

Solve

- \frac{ln ( 2 ) n}{8} = W_{0} (- \frac{ln ( 2 )}{8}) : n = - \frac{8 W _{0} ( - \frac{l n ( 2 )}{8} )}{ln ( 2 )}

n = - \frac{8 W _{- 1} ( - \frac{l n ( 2 )}{8} )}{ln ( 2 )}, n = - \frac{8 W _{0} ( - \frac{l n ( 2 )}{8} )}{ln ( 2 )}

8^{x + 4} = 3

2 5^{2 x + 3} = 2 5^{5 x - 9}

e^{- x} = \frac{1}{3}

(x + 1) e^{x} + e^{x} = 0

672259 = - e^{- 7 \cdot X}

What is the answer to n=2^{(n/8)} ?
The answer to n=2^{(n/8)} is n=-(8\W_{-1}(-(ln(2))/8))/(ln(2)),n=-(8\W_{0}(-(ln(2))/8))/(ln(2))
Solve for n: n=2^{(n/8)}
The solution is n=-(8\W_{-1}(-(ln(2))/8))/(ln(2)),n=-(8\W_{0}(-(ln(2))/8))/(ln(2))