What is the answer to 2^x+2^3x=16 ?

The answer to 2^x+2^3x=16 is x=-(\W_{0}((ln(2))/2)-2ln(2))/(ln(2))

Solve for x: 2^x+2^3x=16

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

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2^{x} + 2^{3} x = 16

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

Decimal Notation

x = 1.61666 \dots

Solution steps

2^{x} + 2^{3} x = 16

Prepare

2^{x} + 2^{3} x = 16

for Lambert form:

1 = (16 - 8 x) e^{- l n (2) x}

Rewrite the equation with

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

and

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

1 = (16 - 8 (- \frac{u - 2 ln ( 2 )}{ln ( 2 )})) e^{- l n (2) (- \frac{u - 2 l n ( 2 )}{l n ( 2 )})}

Simplify

1 = e^{u - 2 l n (2)} (16 + \frac{8 ( u - 2 ln ( 2 ) )}{ln ( 2 )})

Rewrite

1 = e^{u - 2 l n (2)} (16 + \frac{8 ( u - 2 ln ( 2 ) )}{ln ( 2 )})

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 + 2) ln (2),

solve for

x

Solve

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

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

2^{x} = 0.5

m, 2^{n} \cdot n = 2^{m}

5^{2 x - 1} - 10 (x^{5})^{x - 1} - 75 = 0

2^{3 + x} = 5^{x - 4}

2 \cdot 5^{x} + 2 5^{x} = 3

What is the answer to 2^x+2^3x=16 ?
The answer to 2^x+2^3x=16 is x=-(\W_{0}((ln(2))/2)-2ln(2))/(ln(2))
Solve for x: 2^x+2^3x=16
The solution is x=-(\W_{0}((ln(2))/2)-2ln(2))/(ln(2))