What is the answer to log_{4}(x^2+5x+6)=1 ?

The answer to log_{4}(x^2+5x+6)=1 is x=(-5+sqrt(17))/2 ,x=(-5-sqrt(17))/2

Solve for x: log_{4}(x^2+5x+6)=1

The solution is x=(-5+sqrt(17))/2 ,x=(-5-sqrt(17))/2

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lo g_{4} (x^{2} + 5 x + 6) = 1

x = \frac{- 5 + 17}{2}, x = \frac{- 5 - 17}{2}

Decimal

x = - 0.43844 \dots, x = - 4.56155 \dots

Solution steps

lo g_{4} (x^{2} + 5 x + 6) = 1

Apply log rules

x^{2} + 5 x + 6 = 4

Solve

x^{2} + 5 x + 6 = 4 : x = \frac{- 5 + 17}{2}, x = \frac{- 5 - 17}{2}

x = \frac{- 5 + 17}{2}, x = \frac{- 5 - 17}{2}

25 = 5 x + 3 ln (x - 4)

lo g_{x + 1} (9) = 2

lo g_{3} (4 x - 7) = 2

lo g_{10} (2 - 5 x) = 1

lo g_{x} (\frac{1}{27}) = 3

What is the answer to log_{4}(x^2+5x+6)=1 ?
The answer to log_{4}(x^2+5x+6)=1 is x=(-5+sqrt(17))/2 ,x=(-5-sqrt(17))/2
Solve for x: log_{4}(x^2+5x+6)=1
The solution is x=(-5+sqrt(17))/2 ,x=(-5-sqrt(17))/2