What is the answer to (1/2)^x=x ?

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

Solve for x: (1/2)^x=x

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

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(\frac{1}{2})^{x} = x

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

Decimal Notation

x = 0.64118 \dots

Solution steps

(\frac{1}{2})^{x} = x

Prepare

(\frac{1}{2})^{x} = x

for Lambert form:

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

Rewrite the equation with

ln (2) x = u

and

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

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

Rewrite

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

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 = ln (2) x,

solve for

x

Solve

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

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

4^{x - 1} = 16

2^{2 x} + 2^{x + 1} = 80

8^{y} = 32

3^{x} = 15

e^{2 x} = 6 e^{x}

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