What are the solutions to the equation solvefor x,ln(y)=-x^2+c ?

The solutions to the equation solvefor x,ln(y)=-x^2+c are x=sqrt(-ln(y)+c),x=-sqrt(-ln(y)+c)

Find the zeros of solvefor x,ln(y)=-x^2+c

The zeros of solvefor x,ln(y)=-x^2+c are x=sqrt(-ln(y)+c),x=-sqrt(-ln(y)+c)

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solve for

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

x = - ln (y) + c, x = - - ln (y) + c

Solution steps

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

Switch sides

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

Move

c

to the right side

- x^{2} = ln (y) - c

Divide both sides by

- 1

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = - ln (y) + c, x = - - ln (y) + c

4 x^{2} + 8 x + 7

4 x^{2} - 8 x - 13 = 0

8 n^{2} + 8 n + 2 = 0

y, x^{3} - x y^{2} + 2 y^{2} = 0

a^{2} + 2^{2} = 3^{2}

What are the solutions to the equation solvefor x,ln(y)=-x^2+c ?
The solutions to the equation solvefor x,ln(y)=-x^2+c are x=sqrt(-ln(y)+c),x=-sqrt(-ln(y)+c)
Find the zeros of solvefor x,ln(y)=-x^2+c
The zeros of solvefor x,ln(y)=-x^2+c are x=sqrt(-ln(y)+c),x=-sqrt(-ln(y)+c)