What is the answer to 3-ln(x^2-3x+1)= 1/10 ?

The answer to 3-ln(x^2-3x+1)= 1/10 is x=(3+sqrt(4e^{29/10)+5})/2 ,x=(3-sqrt(4e^{29/10)+5})/2

Solve for x: 3-ln(x^2-3x+1)= 1/10

The solution is x=(3+sqrt(4e^{29/10)+5})/2 ,x=(3-sqrt(4e^{29/10)+5})/2

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3 - ln (x^{2} - 3 x + 1) = \frac{1}{10}

x = \frac{3 + 4 e ^{\frac{29}{10}} + 5}{2}, x = \frac{3 - 4 e ^{\frac{29}{10}} + 5}{2}

Solution steps

3 - ln (x^{2} - 3 x + 1) = \frac{1}{10}

Move

3

to the right side

- ln (x^{2} - 3 x + 1) = - \frac{29}{10}

Divide both sides by

- 1

ln (x^{2} - 3 x + 1) = \frac{29}{10}

Apply log rules

x^{2} - 3 x + 1 = e^{\frac{29}{10}}

Solve

x^{2} - 3 x + 1 = e^{\frac{29}{10}} : x = \frac{3 + 4 e ^{\frac{29}{10}} + 5}{2}, x = \frac{3 - 4 e ^{\frac{29}{10}} + 5}{2}

x = \frac{3 + 4 e ^{\frac{29}{10}} + 5}{2}, x = \frac{3 - 4 e ^{\frac{29}{10}} + 5}{2}

Verify Solutions:

x = \frac{3 + 4 e ^{\frac{29}{10}} + 5}{2}

True

, x = \frac{3 - 4 e ^{\frac{29}{10}} + 5}{2}

True

The solutions are

x = \frac{3 + 4 e ^{\frac{29}{10}} + 5}{2}, x = \frac{3 - 4 e ^{\frac{29}{10}} + 5}{2}

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

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

ln (3 x - 16) = 2

ln (3 x - 16) = 3

3 ln (2 x) = - 6

What is the answer to 3-ln(x^2-3x+1)= 1/10 ?
The answer to 3-ln(x^2-3x+1)= 1/10 is x=(3+sqrt(4e^{29/10)+5})/2 ,x=(3-sqrt(4e^{29/10)+5})/2
Solve for x: 3-ln(x^2-3x+1)= 1/10
The solution is x=(3+sqrt(4e^{29/10)+5})/2 ,x=(3-sqrt(4e^{29/10)+5})/2