What is the answer to solvefor x,t(x)=-2^{x-1} ?

The answer to solvefor x,t(x)=-2^{x-1} is x=-(\W(\frac{ln(2))/(2t))}{ln(2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor x,t(x)=-2^{x-1}

solve for

x, t (x) = - 2^{x - 1}

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

Solution steps

t (x) = - 2^{x - 1}

Prepare

t (x) = - 2^{x - 1}

for Lambert form:

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

Rewrite the equation with

- x ln (2) = u

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

e^{u} u = \frac{ln ( 2 )}{2 t} : u = W (\frac{ln ( 2 )}{2 t})

u = W (\frac{ln ( 2 )}{2 t})

Substitute back

u = - x ln (2),

solve for

x

Solve

- x ln (2) = W (\frac{ln ( 2 )}{2 t}) : x = - \frac{W ( \frac{l n ( 2 )}{2 t} )}{ln ( 2 )}

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

2 e^{x} = 3 e^{- x} 5

9^{x} - 1 = 2

e^{4 x} = 7

1 0^{x} = 8

1 0^{x} = e

What is the answer to solvefor x,t(x)=-2^{x-1} ?
The answer to solvefor x,t(x)=-2^{x-1} is x=-(\W(\frac{ln(2))/(2t))}{ln(2)}