What is the answer to ln(432G^2)-2=0 ?

The answer to ln(432G^2)-2=0 is G= e/(12sqrt(3)),G=-e/(12sqrt(3))

Solve for G: ln(432G^2)-2=0

The solution is G= e/(12sqrt(3)),G=-e/(12sqrt(3))

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ln (432 G^{2}) - 2 = 0

G = \frac{e}{12 3}, G = - \frac{e}{12 3}

Decimal Notation

G = 0.13078 \dots, G = - 0.13078 \dots

Solution steps

ln (432 G^{2}) - 2 = 0

Move

2

to the right side

ln (432 G^{2}) = 2

Apply log rules

432 G^{2} = e^{2}

Solve

432 G^{2} = e^{2} : G = \frac{e}{12 3}, G = - \frac{e}{12 3}

G = \frac{e}{12 3}, G = - \frac{e}{12 3}

Verify Solutions:

G = \frac{e}{12 3}

True

, G = - \frac{e}{12 3}

True

The solutions are

G = \frac{e}{12 3}, G = - \frac{e}{12 3}

lo g_{4} (x) = b

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

lo g_{4} (x) = 7

1.482 = - 0.0592 lo g_{10} (x)

2 lo g_{2} (x - 8) + lo g_{2} (8) = 5

What is the answer to ln(432G^2)-2=0 ?
The answer to ln(432G^2)-2=0 is G= e/(12sqrt(3)),G=-e/(12sqrt(3))
Solve for G: ln(432G^2)-2=0
The solution is G= e/(12sqrt(3)),G=-e/(12sqrt(3))