What is the answer to log_{5}(3x^2+7x+3)=0 ?

The answer to log_{5}(3x^2+7x+3)=0 is x=-1/3 ,x=-2

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log_{5}(3x^2+7x+3)=0

lo g_{5} (3 x^{2} + 7 x + 3) = 0

x = - \frac{1}{3}, x = - 2

Decimal Notation

x = - 0.33333 \dots, x = - 2

Solution steps

lo g_{5} (3 x^{2} + 7 x + 3) = 0

Apply log rules

3 x^{2} + 7 x + 3 = 1

Solve

3 x^{2} + 7 x + 3 = 1 : x = - \frac{1}{3}, x = - 2

x = - \frac{1}{3}, x = - 2

ln (x - 1) + ln (x - 2) = ln (7)

1.3648 = 0.6446 ln (x) - 0.5557

3 ln (x) - 2 = 0

c + 5 ln (- 1 + c) = 3.46

lo g_{8} (x + 4) = lo g_{8} (6)

What is the answer to log_{5}(3x^2+7x+3)=0 ?
The answer to log_{5}(3x^2+7x+3)=0 is x=-1/3 ,x=-2
Solve for x: log_{5}(3x^2+7x+3)=0
The solution is x=-1/3 ,x=-2