What is the answer to ln(x^2-x)=1 ?

The answer to ln(x^2-x)=1 is x=(1+sqrt(1+4e))/2 ,x=(1-sqrt(1+4e))/2

Solve for x: ln(x^2-x)=1

The solution is x=(1+sqrt(1+4e))/2 ,x=(1-sqrt(1+4e))/2

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ln (x^{2} - x) = 1

x = \frac{1 + 1 + 4 e}{2}, x = \frac{1 - 1 + 4 e}{2}

Decimal Notation

x = 2.22287 \dots, x = - 1.22287 \dots

Solution steps

ln (x^{2} - x) = 1

Apply log rules

x^{2} - x = e

Solve

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

x = \frac{1 + 1 + 4 e}{2}, x = \frac{1 - 1 + 4 e}{2}

x, 2 = - lo g_{10} (x)

- 1 - 5 \cdot ln (x) = 1

lo g_{2} (k + 2) - lo g_{2} (3 - k) = - 2

lo g_{2} (x) = 1 0^{6}

2 ln (x) + ln (x) = 0

What is the answer to ln(x^2-x)=1 ?
The answer to ln(x^2-x)=1 is x=(1+sqrt(1+4e))/2 ,x=(1-sqrt(1+4e))/2
Solve for x: ln(x^2-x)=1
The solution is x=(1+sqrt(1+4e))/2 ,x=(1-sqrt(1+4e))/2