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

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

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

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

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2 x = 2^{- x}

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

Solution steps

2 x = 2^{- x}

Prepare

2 x = 2^{- x}

for Lambert form:

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

Rewrite the equation with

x ln (2) = u

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

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

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

Substitute back

u = x ln (2),

solve for

x

Solve

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

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

\frac{4 ^{4 x}}{4 ^{x + 2}} = 16

3^{x + 1} = 2^{x - 1}

4^{m + 1} = \frac{1}{8 2}

6 4^{x + 1} = 4^{x + 7}

A = 50 e^{0.07 \cdot t}

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