What is the answer to x16=8^{-x}-4 ?

The answer to x16=8^{-x}-4 is x=(\W_{0}((3ln(2)e^{\frac{3ln(2))/4})/(16))}{3ln(2)}-1/4

Solve for x: x16=8^{-x}-4

The solution is x=(\W_{0}((3ln(2)e^{\frac{3ln(2))/4})/(16))}{3ln(2)}-1/4

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x 16 = 8^{- x} - 4

x = \frac{W _{0} ( \frac{3 l n ( 2 ) e ^{\frac{3 l n ( 2 )}{4}}}{16} )}{3 ln ( 2 )} - \frac{1}{4}

Decimal Notation

x = - 0.16239 \dots

Solution steps

x \cdot 16 = 8^{- x} - 4

Prepare

x 16 = 8^{- x} - 4

for Lambert form:

1 = (x \cdot 16 + 4) e^{3 l n (2) x}

Rewrite the equation with

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

and

x = \frac{u}{3 ln ( 2 )} - \frac{1}{4}

1 = ((\frac{u}{3 ln ( 2 )} - \frac{1}{4}) \cdot 16 + 4) e^{3 l n (2) (\frac{u}{3 l n ( 2 )} - \frac{1}{4})}

Simplify

1 = \frac{16 e ^{u - \frac{3}{4} l n (2)} u}{3 ln ( 2 )}

Rewrite

1 = \frac{16 e ^{u - \frac{3}{4} l n (2)} u}{3 ln ( 2 )}

in Lambert form:

e^{u} u = \frac{3 ln ( 2 ) e ^{\frac{3 l n ( 2 )}{4}}}{16}

Solve

e^{u} u = \frac{3 ln ( 2 ) e ^{\frac{3 l n ( 2 )}{4}}}{16} : u = W_{0} (\frac{3 ln ( 2 ) e ^{\frac{3 l n ( 2 )}{4}}}{16})

u = W_{0} (\frac{3 ln ( 2 ) e ^{\frac{3 l n ( 2 )}{4}}}{16})

Substitute back

u = 3 ln (2) (x + \frac{1}{4}),

solve for

x

Solve

3 ln (2) (x + \frac{1}{4}) = W_{0} (\frac{3 ln ( 2 ) e ^{\frac{3 l n ( 2 )}{4}}}{16}) : x = \frac{W _{0} ( \frac{3 l n ( 2 ) e ^{\frac{3 l n ( 2 )}{4}}}{16} )}{3 ln ( 2 )} - \frac{1}{4}

x = \frac{W _{0} ( \frac{3 l n ( 2 ) e ^{\frac{3 l n ( 2 )}{4}}}{16} )}{3 ln ( 2 )} - \frac{1}{4}

2^{- x} - 4 = 1 1^{6} x

\frac{2 ^{5 x}}{( 2 ^{x} )} = 3 2^{\frac{1}{5}}

n^{n} = 2

e^{x} + 3 e^{- x} = 4

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

What is the answer to x16=8^{-x}-4 ?
The answer to x16=8^{-x}-4 is x=(\W_{0}((3ln(2)e^{\frac{3ln(2))/4})/(16))}{3ln(2)}-1/4
Solve for x: x16=8^{-x}-4
The solution is x=(\W_{0}((3ln(2)e^{\frac{3ln(2))/4})/(16))}{3ln(2)}-1/4