What is the answer to log_{4}(x^2+x+3)=1 ?

The answer to log_{4}(x^2+x+3)=1 is x=(-1+sqrt(5))/2 ,x=(-1-sqrt(5))/2

Solve for x: log_{4}(x^2+x+3)=1

The solution is x=(-1+sqrt(5))/2 ,x=(-1-sqrt(5))/2

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lo g_{4} (x^{2} + x + 3) = 1

x = \frac{- 1 + 5}{2}, x = \frac{- 1 - 5}{2}

Decimal

x = 0.61803 \dots, x = - 1.61803 \dots

Solution steps

lo g_{4} (x^{2} + x + 3) = 1

Apply log rules

x^{2} + x + 3 = 4

Solve

x^{2} + x + 3 = 4 : x = \frac{- 1 + 5}{2}, x = \frac{- 1 - 5}{2}

x = \frac{- 1 + 5}{2}, x = \frac{- 1 - 5}{2}

ln (2 x + 4) = 3

6 lo g_{10} (x - 7) = 13

lo g_{10} (2 x - 3) = 0

2 lo g_{10} (3 x - 2) = 4

lo g_{10} (x + 11) = lo g_{10} (x) + lo g_{10} (11)

What is the answer to log_{4}(x^2+x+3)=1 ?
The answer to log_{4}(x^2+x+3)=1 is x=(-1+sqrt(5))/2 ,x=(-1-sqrt(5))/2
Solve for x: log_{4}(x^2+x+3)=1
The solution is x=(-1+sqrt(5))/2 ,x=(-1-sqrt(5))/2