What is the answer to log_{3}(x^2-3x-1)=0 ?

The answer to log_{3}(x^2-3x-1)=0 is x=(3+sqrt(17))/2 ,x=(3-sqrt(17))/2

Solve for x: log_{3}(x^2-3x-1)=0

The solution is x=(3+sqrt(17))/2 ,x=(3-sqrt(17))/2

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lo g_{3} (x^{2} - 3 x - 1) = 0

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

Decimal Notation

x = 3.56155 \dots, x = - 0.56155 \dots

Solution steps

lo g_{3} (x^{2} - 3 x - 1) = 0

Apply log rules

x^{2} - 3 x - 1 = 1

Solve

x^{2} - 3 x - 1 = 1 : x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

x = \frac{3 + 17}{2}, x = \frac{3 - 17}{2}

lo g_{2} (x - 5) = 3 - lo g_{2} (x - 7)

ln (x) + ln (x - 1) = ln (6)

lo g_{16} (x) = (lo g_{4} (x))^{2}

lo g_{10} (x) = 7

ln (y) = 0

What is the answer to log_{3}(x^2-3x-1)=0 ?
The answer to log_{3}(x^2-3x-1)=0 is x=(3+sqrt(17))/2 ,x=(3-sqrt(17))/2
Solve for x: log_{3}(x^2-3x-1)=0
The solution is x=(3+sqrt(17))/2 ,x=(3-sqrt(17))/2