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))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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 )}

208 (1.04)^{t} = 279 (1.01)^{t}

15 = A (1 - e^{- k^{2}})

36 = 18 (1 6^{0.0125 x})

1280 = 5 (2^{t})

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

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))