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)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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 )}

5 e^{2 - x} = 1 0^{x + 4}

- (16 x - 16) e^{- x} = 0

2^{x + 1} + 6 (2^{x - 1}) = 12

\frac{1}{36} = 6^{x - 3}

2^{y} = \frac{1}{8}

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)}