What is the answer to log_{4}(3x^2+6x+7)=2 ?

The answer to log_{4}(3x^2+6x+7)=2 is x=1,x=-3

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log_{4}(3x^2+6x+7)=2

lo g_{4} (3 x^{2} + 6 x + 7) = 2

x = 1, x = - 3

Solution steps

lo g_{4} (3 x^{2} + 6 x + 7) = 2

Apply log rules

3 x^{2} + 6 x + 7 = 16

Solve

3 x^{2} + 6 x + 7 = 16 : x = 1, x = - 3

x = 1, x = - 3

ln (ln (5 x)) = 0

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

0.49 = ln (x)

lo g_{b} (25) = 2

lo g_{10} (x^{2} + 6 x + 10) = 0

What is the answer to log_{4}(3x^2+6x+7)=2 ?
The answer to log_{4}(3x^2+6x+7)=2 is x=1,x=-3
Solve for x: log_{4}(3x^2+6x+7)=2
The solution is x=1,x=-3