What is the answer to x=2^{x/8} ?

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

Solve for x: x=2^{x/8}

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

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

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

Decimal Notation

x = 43.55926 \dots, x = 1.09999 \dots

Solution steps

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

Prepare

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

for Lambert form:

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

Rewrite the equation with

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

and

x = - \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 ) x}{8},

solve for

x

Solve

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

Solve

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

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

0.8 = - e^{\frac{- 2 t - ( - 2 )}{7}} + 1

6 = 2^{4 x - 3}

z, e^{2 x + 2 y + z} = 1

x, 4^{2 x^{2} + 2 x} = 8

12 5^{r} = 5

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