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

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

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

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

x = 3, x = - 3

Solution steps

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

Apply log rules

x^{2} - 3 x + 1 = - 3 x + 10

Solve

x^{2} - 3 x + 1 = - 3 x + 10 : x = 3, x = - 3

x = 3, x = - 3

Verify Solutions:

x = 3

True

, x = - 3

True

The solutions are

x = 3, x = - 3

x, lo g_{6} (x) + lo g_{6} (y) = 1

lo g_{2} (x - 4) = 0

lo g_{3} (2 x) = 2

a, lo g_{3} (a) - 2 \cdot lo g_{10} (b) = 3

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

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