What is the answer to 0=ln|y^2-y| ?

The answer to 0=ln|y^2-y| is y=(1-sqrt(5))/2 ,y=(1+sqrt(5))/2

Solve for y: 0=ln|y^2-y|

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

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0 = ln y^{2} - y

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

Solution steps

0 = ln y^{2} - y

Apply log rules

1 = y^{2} - y

Solve

1 = y^{2} - y : y = \frac{1 - 5}{2} or y = \frac{1 + 5}{2}

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

(11 i) (- 4 i)

5 \cdot 45

3 x^{2} - 8 x + 4 > 0

\frac{1}{3} (x - 4) = \frac{1}{2} (x - 6)

\frac{1}{1 + i} \cdot \frac{1 - i}{1 - i}

What is the answer to 0=ln|y^2-y| ?
The answer to 0=ln|y^2-y| is y=(1-sqrt(5))/2 ,y=(1+sqrt(5))/2
Solve for y: 0=ln|y^2-y|
The solution is y=(1-sqrt(5))/2 ,y=(1+sqrt(5))/2