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

The answer to solvefor x,y=ln(2x^2-3y^2) is x=sqrt((e^y+3y^2)/2),x=-sqrt((e^y+3y^2)/2)

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

solve for

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

x = \frac{e ^{y} + 3 y ^{2}}{2}, x = - \frac{e ^{y} + 3 y ^{2}}{2}

Solution steps

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

Apply log rules

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

Solve

e^{y} = 2 x^{2} - 3 y^{2} : x = \frac{e ^{y} + 3 y ^{2}}{2}, x = - \frac{e ^{y} + 3 y ^{2}}{2}

x = \frac{e ^{y} + 3 y ^{2}}{2}, x = - \frac{e ^{y} + 3 y ^{2}}{2}

lo g_{2 x} (3 x^{2} + 9 x - 20) = 2

4 = lo g_{x} (625)

7 x^{6} ln (x) + x^{6} = 0

lo g_{3} (6 - x) = 3

lo g_{2} (3 x + 1) = 3

What is the answer to solvefor x,y=ln(2x^2-3y^2) ?
The answer to solvefor x,y=ln(2x^2-3y^2) is x=sqrt((e^y+3y^2)/2),x=-sqrt((e^y+3y^2)/2)