What is the answer to solvefor x,2y^2+3y-5x=ln(x+y^2)-1 ?

The answer to solvefor x,2y^2+3y-5x=ln(x+y^2)-1 is x=(\W(5/(e^{-7y^2-3y-1)})-5y^2)/5

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solvefor x,2y^2+3y-5x=ln(x+y^2)-1

solve for

x, 2 y^{2} + 3 y - 5 x = ln (x + y^{2}) - 1

x = \frac{W ( \frac{5}{e ^{- 7 y^{2} - 3 y - 1}} ) - 5 y ^{2}}{5}

Solution steps

2 y^{2} + 3 y - 5 x = ln (x + y^{2}) - 1

Switch sides

ln (x + y^{2}) - 1 = 2 y^{2} + 3 y - 5 x

Move

1

to the right side

ln (x + y^{2}) = 2 y^{2} + 3 y - 5 x + 1

f (x) = g (x),

then

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

e^{l n (x + y^{2})} = e^{2 y^{2} + 3 y - 5 x + 1}

Simplify

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

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

Solve

x + y^{2} = e^{2 y^{2} + 3 y - 5 x + 1} : x = \frac{W ( \frac{5}{e ^{- 7 y^{2} - 3 y - 1}} ) - 5 y ^{2}}{5}

x = \frac{W ( \frac{5}{e ^{- 7 y^{2} - 3 y - 1}} ) - 5 y ^{2}}{5}

Verify Solutions:

x = \frac{W ( \frac{5}{e ^{- 7 y^{2} - 3 y - 1}} ) - 5 y ^{2}}{5}

True

The solution is

x = \frac{W ( \frac{5}{e ^{- 7 y^{2} - 3 y - 1}} ) - 5 y ^{2}}{5}

lo g_{7} (x) = lo g_{7} (a) \cdot b^{2} - lo g_{7} (a^{3} \cdot b)

- 4 = 20 \cdot lo g_{10} (\frac{x}{80})

ln (x^{2} + 6) - ln (x - 2) = ln (x + 1)

lo g_{10} (n) = 1.0043

lo g_{10} (x) + lo g_{10} (7) = lo g_{10} (4)

What is the answer to solvefor x,2y^2+3y-5x=ln(x+y^2)-1 ?
The answer to solvefor x,2y^2+3y-5x=ln(x+y^2)-1 is x=(\W(5/(e^{-7y^2-3y-1)})-5y^2)/5