What is the answer to ln(12x^2+12)-3=0 ?

The answer to ln(12x^2+12)-3=0 is x=(sqrt(e^3-12))/(2sqrt(3)),x=-(sqrt(e^3-12))/(2sqrt(3))

Solve for x: ln(12x^2+12)-3=0

The solution is x=(sqrt(e^3-12))/(2sqrt(3)),x=-(sqrt(e^3-12))/(2sqrt(3))

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ln (12 x^{2} + 12) - 3 = 0

x = \frac{e ^{3} - 12}{2 3}, x = - \frac{e ^{3} - 12}{2 3}

Decimal

x = 0.82085 \dots, x = - 0.82085 \dots

Solution steps

ln (12 x^{2} + 12) - 3 = 0

Move

3

to the right side

ln (12 x^{2} + 12) = 3

Apply log rules

12 x^{2} + 12 = e^{3}

Solve

12 x^{2} + 12 = e^{3} : x = \frac{e ^{3} - 12}{2 3}, x = - \frac{e ^{3} - 12}{2 3}

x = \frac{e ^{3} - 12}{2 3}, x = - \frac{e ^{3} - 12}{2 3}

Verify Solutions:

x = \frac{e ^{3} - 12}{2 3}

True

, x = - \frac{e ^{3} - 12}{2 3}

True

The solutions are

x = \frac{e ^{3} - 12}{2 3}, x = - \frac{e ^{3} - 12}{2 3}

2 x ln (x) + x

lo g_{2} (x^{2} + x + 2) = 2

lo g_{x} (\frac{1}{3125}) = - 5

lo g_{9} (x) + lo g_{9} (x + 2) - \frac{1}{2} = 0

lo g_{a} (2) = 1

What is the answer to ln(12x^2+12)-3=0 ?
The answer to ln(12x^2+12)-3=0 is x=(sqrt(e^3-12))/(2sqrt(3)),x=-(sqrt(e^3-12))/(2sqrt(3))
Solve for x: ln(12x^2+12)-3=0
The solution is x=(sqrt(e^3-12))/(2sqrt(3)),x=-(sqrt(e^3-12))/(2sqrt(3))