What is the answer to log_{10}(x^2-3x+2)=1 ?

The answer to log_{10}(x^2-3x+2)=1 is x=(3+sqrt(41))/2 ,x=(3-sqrt(41))/2

Solve for x: log_{10}(x^2-3x+2)=1

The solution is x=(3+sqrt(41))/2 ,x=(3-sqrt(41))/2

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lo g_{10} (x^{2} - 3 x + 2) = 1

x = \frac{3 + 41}{2}, x = \frac{3 - 41}{2}

Decimal Notation

x = 4.70156 \dots, x = - 1.70156 \dots

Solution steps

lo g_{10} (x^{2} - 3 x + 2) = 1

Apply log rules

x^{2} - 3 x + 2 = 10

Solve

x^{2} - 3 x + 2 = 10 : x = \frac{3 + 41}{2}, x = \frac{3 - 41}{2}

x = \frac{3 + 41}{2}, x = \frac{3 - 41}{2}

lo g_{a} (16) = \frac{4}{5}

ln (x) = \frac{- 1}{30}

lo g_{5} (x^{2} + 8) = 3

\frac{n}{lo g _{10} ( n )} = 8

1.25 = lo g_{10} (x)

What is the answer to log_{10}(x^2-3x+2)=1 ?
The answer to log_{10}(x^2-3x+2)=1 is x=(3+sqrt(41))/2 ,x=(3-sqrt(41))/2
Solve for x: log_{10}(x^2-3x+2)=1
The solution is x=(3+sqrt(41))/2 ,x=(3-sqrt(41))/2