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

The answer to solvefor x,y=sqrt(ln(x^2)) is x=sqrt(e^{y^2)} {y>= 0},x=-sqrt(e^{y^2)} {y>= 0}

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solvefor x,y=sqrt(ln(x^2))

solve for

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

x = e^{y^{2}} {y \geq 0}, x = - e^{y^{2}} {y \geq 0}

Solution steps

y = ln (x^{2})

Rewrite the equation with

ln (x^{2}) = u

y = u

Solve

y = u : u = y^{2} {y \geq 0}

u = y^{2} {y \geq 0}

Substitute back

u = ln (x^{2}),

solve for

x

Solve

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

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

Verify Solutions:

x = e^{y^{2}} {y \geq 0}, x = - e^{y^{2}} {y \geq 0}

The solutions are

x = e^{y^{2}} {y \geq 0}, x = - e^{y^{2}} {y \geq 0}

0.789 = ln (x^{3})

p \cdot ln (A) = ln (A^{p})

3 ln (2 x) = 6

x, 5 + lo g_{2} (- 5 x) = 9

3 + lo g_{8} (- 5 n) = 6

What is the answer to solvefor x,y=sqrt(ln(x^2)) ?
The answer to solvefor x,y=sqrt(ln(x^2)) is x=sqrt(e^{y^2)} {y>= 0},x=-sqrt(e^{y^2)} {y>= 0}