What is the answer to x-2^{-x}=0 ?

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

Solve for x: x-2^{-x}=0

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

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x - 2^{- x} = 0

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

Solution steps

x - 2^{- x} = 0

Prepare

x - 2^{- x} = 0

for Lambert form:

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

Rewrite the equation with

x ln (2) = u

and

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

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

Rewrite

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

in Lambert form:

e^{u} u = ln (2)

Solve

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

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

Substitute back

u = x ln (2),

solve for

x

Solve

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

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

x, e^{x y} + y = x - 1

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

4^{x} - 2^{x + 1} = 48

(7^{- x})^{2} = 343

5^{x + 2} = 4^{x}

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