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

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

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

solve for

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

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

Solution steps

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

Apply log rules

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

Solve

e^{z} = x^{2} + y^{2} : x = e^{z} - y^{2}, x = - e^{z} - y^{2}

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

lo g_{10} (x^{2} + 3 x) = 1

lo g_{32} (x) = \frac{1}{5}

0.345 = ln (x^{2})

lo g_{10} (x) + lo g_{10} (x + 1) = 1

lo g_{7} (x + 1) + lo g_{7} (x - 5) = 1

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