What is the answer to log_{2}((2x^2+7x)+9)=3 ?

The answer to log_{2}((2x^2+7x)+9)=3 is x=(-7+sqrt(41))/4 ,x=(-7-sqrt(41))/4

Solve for x: log_{2}((2x^2+7x)+9)=3

The solution is x=(-7+sqrt(41))/4 ,x=(-7-sqrt(41))/4

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lo g_{2} ((2 x^{2} + 7 x) + 9) = 3

x = \frac{- 7 + 41}{4}, x = \frac{- 7 - 41}{4}

Decimal

x = - 0.14921 \dots, x = - 3.35078 \dots

Solution steps

lo g_{2} ((2 x^{2} + 7 x) + 9) = 3

Apply log rules

2 x^{2} + 7 x + 9 = 8

Solve

2 x^{2} + 7 x + 9 = 8 : x = \frac{- 7 + 41}{4}, x = \frac{- 7 - 41}{4}

x = \frac{- 7 + 41}{4}, x = \frac{- 7 - 41}{4}

lo g_{5} (5 x) + lo g_{5} (x) = 0

lo g_{2} (x) = 12

lo g_{4} (x^{2} + 15) = 3

\frac{- 3 ln ( x ) + 1}{x ^{4}} = 0

lo g_{3} (12 - 5 x) = 2

What is the answer to log_{2}((2x^2+7x)+9)=3 ?
The answer to log_{2}((2x^2+7x)+9)=3 is x=(-7+sqrt(41))/4 ,x=(-7-sqrt(41))/4
Solve for x: log_{2}((2x^2+7x)+9)=3
The solution is x=(-7+sqrt(41))/4 ,x=(-7-sqrt(41))/4