What is the answer to solvefor y,d(y)=7(2)^y ?

The answer to solvefor y,d(y)=7(2)^y is y=-(\W(-(7ln(2))/d))/(ln(2))

Popular >

solvefor y,d(y)=7(2)^y

solve for

y, d (y) = 7 (2)^{y}

y = - \frac{W ( - \frac{7 l n ( 2 )}{d} )}{ln ( 2 )}

Solution steps

d (y) = 7 (2)^{y}

Prepare

d (y) = 7 (2)^{y}

for Lambert form:

d y e^{- l n (2) y} = 7

Rewrite the equation with

- y ln (2) = u

and

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

d (- \frac{u}{ln ( 2 )}) e^{u} = 7

Rewrite

d (- \frac{u}{ln ( 2 )}) e^{u} = 7

in Lambert form:

e^{u} u = - \frac{7 ln ( 2 )}{d}

Solve

e^{u} u = - \frac{7 ln ( 2 )}{d} : u = W (- \frac{7 ln ( 2 )}{d})

u = W (- \frac{7 ln ( 2 )}{d})

Substitute back

u = - y ln (2),

solve for

y

Solve

- y ln (2) = W (- \frac{7 ln ( 2 )}{d}) : y = - \frac{W ( - \frac{7 l n ( 2 )}{d} )}{ln ( 2 )}

y = - \frac{W ( - \frac{7 l n ( 2 )}{d} )}{ln ( 2 )}

y^{^{''}} + 4 y^{^{'}} + 5 y = 1 - 5 x + 8 (cos (x) - sin (x))

4 x^{^{''}} - 4 x^{^{'}} + x = \frac{8}{t ^{2}} e^{\frac{t}{2}}

- y^{^{''}} + 2 y^{^{'}} + 3 y = 1

y^{^{''}} - y^{^{'}} = e^{x} (3 sin (x) + cos (x))

4 \frac{d ^{2} y}{d x ^{2}} + 9 y = 15

What is the answer to solvefor y,d(y)=7(2)^y ?
The answer to solvefor y,d(y)=7(2)^y is y=-(\W(-(7ln(2))/d))/(ln(2))