What is the answer to ln(y^2)=ln(y+3)+2 ?

The answer to ln(y^2)=ln(y+3)+2 is y=(e^2+sqrt(e^4+12e^2))/2 ,y=(e^2-sqrt(e^4+12e^2))/2

Solve for y: ln(y^2)=ln(y+3)+2

The solution is y=(e^2+sqrt(e^4+12e^2))/2 ,y=(e^2-sqrt(e^4+12e^2))/2

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

y = \frac{e ^{2} + e ^{4} + 12 e ^{2}}{2}, y = \frac{e ^{2} - e ^{4} + 12 e ^{2}}{2}

Decimal Notation

y = 9.67923 \dots, y = - 2.29017 \dots

Solution steps

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

Apply log rules

y^{2} = (y + 3) e^{2}

Solve

y^{2} = (y + 3) e^{2} : y = \frac{e ^{2} + e ^{4} + 12 e ^{2}}{2}, y = \frac{e ^{2} - e ^{4} + 12 e ^{2}}{2}

y = \frac{e ^{2} + e ^{4} + 12 e ^{2}}{2}, y = \frac{e ^{2} - e ^{4} + 12 e ^{2}}{2}

Verify Solutions:

y = \frac{e ^{2} + e ^{4} + 12 e ^{2}}{2}

True

, y = \frac{e ^{2} - e ^{4} + 12 e ^{2}}{2}

True

The solutions are

y = \frac{e ^{2} + e ^{4} + 12 e ^{2}}{2}, y = \frac{e ^{2} - e ^{4} + 12 e ^{2}}{2}

x, y = ln (3 x)

x^{2} ln (x) - 9 ln (x) = 0

x, ln (x) = z

lo g_{5} (x + 1) = 2 + lo g_{5} (3 x - 2)

ln ∣ x ∣ = ln (x^{2})

What is the answer to ln(y^2)=ln(y+3)+2 ?
The answer to ln(y^2)=ln(y+3)+2 is y=(e^2+sqrt(e^4+12e^2))/2 ,y=(e^2-sqrt(e^4+12e^2))/2
Solve for y: ln(y^2)=ln(y+3)+2
The solution is y=(e^2+sqrt(e^4+12e^2))/2 ,y=(e^2-sqrt(e^4+12e^2))/2