What is the answer to log_{7}(x^2-3x+3)=0 ?

The answer to log_{7}(x^2-3x+3)=0 is x=2,x=1

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log_{7}(x^2-3x+3)=0

lo g_{7} (x^{2} - 3 x + 3) = 0

x = 2, x = 1

Solution steps

lo g_{7} (x^{2} - 3 x + 3) = 0

Apply log rules

x^{2} - 3 x + 3 = 1

Solve

x^{2} - 3 x + 3 = 1 : x = 2, x = 1

x = 2, x = 1

lo g_{3} (7 x - 15) = 3

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

lo g_{4} (\frac{X + 3}{4}) = - 1

lo g_{x} (\frac{27}{8}) = 3

lo g_{4 x} (3) = 2

What is the answer to log_{7}(x^2-3x+3)=0 ?
The answer to log_{7}(x^2-3x+3)=0 is x=2,x=1
Solve for x: log_{7}(x^2-3x+3)=0
The solution is x=2,x=1