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

The answer to ln(m^2-m+1)=1 is m=(1+sqrt(4e-3))/2 ,m=(1-sqrt(4e-3))/2

Solve for m: ln(m^2-m+1)=1

The solution is m=(1+sqrt(4e-3))/2 ,m=(1-sqrt(4e-3))/2

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ln (m^{2} - m + 1) = 1

m = \frac{1 + 4 e - 3}{2}, m = \frac{1 - 4 e - 3}{2}

Decimal Notation

m = 1.90295 \dots, m = - 0.90295 \dots

Solution steps

ln (m^{2} - m + 1) = 1

Apply log rules

m^{2} - m + 1 = e

Solve

m^{2} - m + 1 = e : m = \frac{1 + 4 e - 3}{2}, m = \frac{1 - 4 e - 3}{2}

m = \frac{1 + 4 e - 3}{2}, m = \frac{1 - 4 e - 3}{2}

lo g_{\frac{1}{6}} (x) = - 2

3 \cdot ln (t) = 5

x, lo g_{x} (2 x + 15) = 2

lo g_{10} (n) = 3600000

What is the answer to ln(m^2-m+1)=1 ?
The answer to ln(m^2-m+1)=1 is m=(1+sqrt(4e-3))/2 ,m=(1-sqrt(4e-3))/2
Solve for m: ln(m^2-m+1)=1
The solution is m=(1+sqrt(4e-3))/2 ,m=(1-sqrt(4e-3))/2