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

The answer to solvefor x,y^2+ln(xy)=2 is x=(e^{2-y^2})/y {y 0}

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

solve for

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

x = \frac{e ^{2 - y^{2}}}{y} {y < 0 or y > 0}

Solution steps

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

Move

y^{2}

to the right side

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

Apply log rules

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

Solve

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

x = \frac{e ^{2 - y^{2}}}{y}

Verify Solutions:

x = \frac{e ^{2 - y^{2}}}{y} {y < 0 or y > 0}

The solution is

x = \frac{e ^{2 - y^{2}}}{y} {y < 0 or y > 0}

3 lo g_{2} (x) = 5

ln (x^{8}) - 9 ln (x) = 6

- 8 lo g_{10} (b) = - 64

3 lo g_{2} (x^{2} + 6 x) + 5 = 17

lo g_{2} (x^{2} - 7 x) = 3

What is the answer to solvefor x,y^2+ln(xy)=2 ?
The answer to solvefor x,y^2+ln(xy)=2 is x=(e^{2-y^2})/y {y<0\lor y>0}