What is the answer to solvefor y,x=2y+4^y ?

The answer to solvefor y,x=2y+4^y is y=-(\W(ln(2)*2^x)-ln(2)x)/(2ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor y,x=2y+4^y

solve for

y, x = 2 y + 4^{y}

y = - \frac{W ( ln ( 2 ) \cdot 2 ^{x} ) - ln ( 2 ) x}{2 ln ( 2 )}

Solution steps

x = 2 y + 4^{y}

Prepare

x = 2 y + 4^{y}

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

e^{u} u = ln (2) \cdot 2^{x}

Solve

e^{u} u = ln (2) \cdot 2^{x} : u = W (ln (2) \cdot 2^{x})

u = W (ln (2) \cdot 2^{x})

Substitute back

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

solve for

y

Solve

2 ln (2) (- y + \frac{1}{2} x) = W (ln (2) \cdot 2^{x}) : y = - \frac{W ( ln ( 2 ) \cdot 2 ^{x} ) - ln ( 2 ) x}{2 ln ( 2 )}

y = - \frac{W ( ln ( 2 ) \cdot 2 ^{x} ) - ln ( 2 ) x}{2 ln ( 2 )}

2^{x + 1} = 3^{x - 5}

2. 2^{x} = 8

3^{x^{2}} = 9^{- x + 12}

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

4^{2 x} - 68 \cdot 4^{x} + 256 = 0

What is the answer to solvefor y,x=2y+4^y ?
The answer to solvefor y,x=2y+4^y is y=-(\W(ln(2)*2^x)-ln(2)x)/(2ln(2))