What is the answer to solvefor x,e^{x^2+y^2}=3 ?

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

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solvefor x,e^{x^2+y^2}=3

solve for

x, e^{x^{2} + y^{2}} = 3

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

Solution steps

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

Apply exponent rules

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

Solve

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

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

\frac{e ^{x} - 1}{x} = 1

3^{x} - 9^{2 x} = 2 7^{3 x}

7^{2 x - 6} = 1

3 6^{x} = \frac{1}{6}

9^{x + 1} = 2 7^{2 x - 3}

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