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

The answer to log_{2}(3x^2+7x)=3 is x=(-7+sqrt(145))/6 ,x=(-7-sqrt(145))/6

Solve for x: log_{2}(3x^2+7x)=3

The solution is x=(-7+sqrt(145))/6 ,x=(-7-sqrt(145))/6

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

x = \frac{- 7 + 145}{6}, x = \frac{- 7 - 145}{6}

Decimal Notation

x = 0.84026 \dots, x = - 3.17359 \dots

Solution steps

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

Apply log rules

3 x^{2} + 7 x = 8

Solve

3 x^{2} + 7 x = 8 : x = \frac{- 7 + 145}{6}, x = \frac{- 7 - 145}{6}

x = \frac{- 7 + 145}{6}, x = \frac{- 7 - 145}{6}

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

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

2 lo g_{10} (x) - lo g_{10} (2 x + 15) = 0

lo g_{4} (m) = 0

lo g_{4} (x) = 0

What is the answer to log_{2}(3x^2+7x)=3 ?
The answer to log_{2}(3x^2+7x)=3 is x=(-7+sqrt(145))/6 ,x=(-7-sqrt(145))/6
Solve for x: log_{2}(3x^2+7x)=3
The solution is x=(-7+sqrt(145))/6 ,x=(-7-sqrt(145))/6