What is the answer to solvefor x,y=ln(xy^2)+x-xy+1 ?

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

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solvefor x,y=ln(xy^2)+x-xy+1

solve for

x, y = ln (x y^{2}) + x - x y + 1

x = \frac{W ( \frac{- y + 1}{e ^{- y + 1} y ^{2}} )}{- y + 1}

Solution steps

y = ln (x y^{2}) + x - x y + 1

Switch sides

ln (x y^{2}) + x - x y + 1 = y

Move

x

to the right side

ln (x y^{2}) - x y + 1 = y - x

Move

x y

to the right side

ln (x y^{2}) + 1 = y - x + x y

Move

1

to the right side

ln (x y^{2}) = y - x + x y - 1

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

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

Simplify

e^{l n (x y^{2})} : x y^{2}

x y^{2} = e^{y - x + x y - 1}

Solve

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

x = \frac{W ( \frac{- y + 1}{e ^{- y + 1} y ^{2}} )}{- y + 1}

Verify Solutions:

x = \frac{W ( \frac{- y + 1}{e ^{- y + 1} y ^{2}} )}{- y + 1}

True

The solution is

x = \frac{W ( \frac{- y + 1}{e ^{- y + 1} y ^{2}} )}{- y + 1}

lo g_{3 x} (x^{2} + 4 x + 15) = 2

lo g_{2} (p) = 8

0 = 3 x^{2} ln (x) + x^{2}

3 - lo g_{3} (x) - 2 = 0

lo g_{10} (x) (32) = \frac{5}{2}

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