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

The answer to 0=-(ln(-3(0)^2+c))/2 is c=1

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0=-(ln(-3(0)^2+c))/2

0 = - \frac{ln ( - 3 ( 0 ) ^{2} + c )}{2}

c = 1

Solution steps

0 = - \frac{ln ( - 3 ( 0 ) ^{2} + c )}{2}

Switch sides

- \frac{ln ( - 3 \cdot 0 ^{2} + c )}{2} = 0

Multiply both sides by

2

- ln (- 3 \cdot 0^{2} + c) = 0

Divide both sides by

- 1

ln (- 3 \cdot 0^{2} + c) = 0

Apply log rules

c = 1

Verify Solutions:

c = 1

True

The solution is

c = 1

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

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

lo g_{9} (2 x - 5) = 0.5

lo g_{9} (27 x) + lo g_{x} (81) = 6

lo g_{10} (x) 27 = - 3

What is the answer to 0=-(ln(-3(0)^2+c))/2 ?
The answer to 0=-(ln(-3(0)^2+c))/2 is c=1
Solve for c: 0=-(ln(-3(0)^2+c))/2
The solution is c=1