What is the answer to (4^x)^2+4x=2^{-6} ?

The answer to (4^x)^2+4x=2^{-6} is x=-(64\W_{0}(e^{\frac{ln(2))/(64)}ln(2))-ln(2)}{256ln(2)}

Solve for x: (4^x)^2+4x=2^{-6}

The solution is x=-(64\W_{0}(e^{\frac{ln(2))/(64)}ln(2))-ln(2)}{256ln(2)}

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(4^{x})^{2} + 4 x = 2^{- 6}

x = - \frac{64 W _{0} ( e ^{\frac{l n ( 2 )}{64}} ln ( 2 ) ) - ln ( 2 )}{256 ln ( 2 )}

Decimal Notation

x = - 0.15759 \dots

Solution steps

(4^{x})^{2} + 4 x = 2^{- 6}

Prepare

(4^{x})^{2} + 4 x = 2^{- 6}

for Lambert form:

1 = (\frac{1}{64} - 4 x) e^{- 4 l n (2) x}

Rewrite the equation with

4 ln (2) (- x + \frac{1}{256}) = u

and

x = - \frac{64 u - ln ( 2 )}{256 ln ( 2 )}

1 = (\frac{1}{64} - 4 (- \frac{64 u - ln ( 2 )}{256 ln ( 2 )})) e^{- 4 l n (2) (- \frac{64 u - l n ( 2 )}{256 l n ( 2 )})}

Simplify

1 = e^{\frac{64 u - l n ( 2 )}{64}} (\frac{1}{64} + \frac{64 u - ln ( 2 )}{64 ln ( 2 )})

Rewrite

1 = e^{\frac{64 u - l n ( 2 )}{64}} (\frac{1}{64} + \frac{64 u - ln ( 2 )}{64 ln ( 2 )})

in Lambert form:

e^{u} u = e^{\frac{l n ( 2 )}{64}} ln (2)

Solve

e^{u} u = e^{\frac{l n ( 2 )}{64}} ln (2) : u = W_{0} (e^{\frac{l n ( 2 )}{64}} ln (2))

u = W_{0} (e^{\frac{l n ( 2 )}{64}} ln (2))

Substitute back

u = 4 ln (2) (- x + \frac{1}{256}),

solve for

x

Solve

4 ln (2) (- x + \frac{1}{256}) = W_{0} (e^{\frac{l n ( 2 )}{64}} ln (2)) : x = - \frac{64 W _{0} ( e ^{\frac{l n ( 2 )}{64}} ln ( 2 ) ) - ln ( 2 )}{256 ln ( 2 )}

x = - \frac{64 W _{0} ( e ^{\frac{l n ( 2 )}{64}} ln ( 2 ) ) - ln ( 2 )}{256 ln ( 2 )}

e^{7 x + 7} = 7 e^{7 - x}

(\frac{2}{3})^{x} = 2^{x - 1}

4^{x} - 4^{\frac{( 2 x + 3 )}{2}} = - 7

27 (3^{2 x}) + 80 (3^{x}) - 3 = 0

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

What is the answer to (4^x)^2+4x=2^{-6} ?
The answer to (4^x)^2+4x=2^{-6} is x=-(64\W_{0}(e^{\frac{ln(2))/(64)}ln(2))-ln(2)}{256ln(2)}
Solve for x: (4^x)^2+4x=2^{-6}
The solution is x=-(64\W_{0}(e^{\frac{ln(2))/(64)}ln(2))-ln(2)}{256ln(2)}