What is the answer to 2^{(x)}+4x+12=0 ?

The answer to 2^{(x)}+4x+12=0 is x=-(\W_{0}(\frac{ln(2))/(32))+3ln(2)}{ln(2)}

Solve for x: 2^{(x)}+4x+12=0

The solution is x=-(\W_{0}(\frac{ln(2))/(32))+3ln(2)}{ln(2)}

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2^{(x)} + 4 x + 12 = 0

x = - \frac{W _{0} ( \frac{l n ( 2 )}{32} ) + 3 ln ( 2 )}{ln ( 2 )}

Decimal Notation

x = - 3.03059 \dots

Solution steps

2^{(x)} + 4 x + 12 = 0

Prepare

2^{(x)} + 4 x + 12 = 0

for Lambert form:

1 = - e^{- l n (2) x} (4 x + 12)

Rewrite the equation with

(- x - 3) ln (2) = u

and

x = - \frac{u + 3 ln ( 2 )}{ln ( 2 )}

1 = - e^{- l n (2) (- \frac{u + 3 l n ( 2 )}{l n ( 2 )})} (4 (- \frac{u + 3 ln ( 2 )}{ln ( 2 )}) + 12)

Simplify

1 = - e^{u + 3 l n (2)} (- \frac{4 ( u + 3 ln ( 2 ) )}{ln ( 2 )} + 12)

Rewrite

1 = - e^{u + 3 l n (2)} (- \frac{4 ( u + 3 ln ( 2 ) )}{ln ( 2 )} + 12)

in Lambert form:

e^{u} u = \frac{ln ( 2 )}{32}

Solve

e^{u} u = \frac{ln ( 2 )}{32} : u = W_{0} (\frac{ln ( 2 )}{32})

u = W_{0} (\frac{ln ( 2 )}{32})

Substitute back

u = (- x - 3) ln (2),

solve for

x

Solve

(- x - 3) ln (2) = W_{0} (\frac{ln ( 2 )}{32}) : x = - \frac{W _{0} ( \frac{l n ( 2 )}{32} ) + 3 ln ( 2 )}{ln ( 2 )}

x = - \frac{W _{0} ( \frac{l n ( 2 )}{32} ) + 3 ln ( 2 )}{ln ( 2 )}

e^{- x} + 3 = 0

3 6^{\frac{1}{2} - 4 x} = 6

2^{5} + 2^{x - 1} = 2^{x}

3^{2 - x} + 3^{3 - x} = 12

(- 5)^{x} = - 125

What is the answer to 2^{(x)}+4x+12=0 ?
The answer to 2^{(x)}+4x+12=0 is x=-(\W_{0}(\frac{ln(2))/(32))+3ln(2)}{ln(2)}
Solve for x: 2^{(x)}+4x+12=0
The solution is x=-(\W_{0}(\frac{ln(2))/(32))+3ln(2)}{ln(2)}