What is the answer to log_{3}(x^2-2x+1)=log_{3}(x^2+x+7) ?

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

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log_{3}(x^2-2x+1)=log_{3}(x^2+x+7)

lo g_{3} (x^{2} - 2 x + 1) = lo g_{3} (x^{2} + x + 7)

x = - 2

Solution steps

lo g_{3} (x^{2} - 2 x + 1) = lo g_{3} (x^{2} + x + 7)

Apply log rules

x^{2} - 2 x + 1 = x^{2} + x + 7

Solve

x^{2} - 2 x + 1 = x^{2} + x + 7 : x = - 2

x = - 2

Verify Solutions:

x = - 2

True

The solution is

x = - 2

x, ln (\frac{x}{y}) = 3

13 lo g_{10} (10 x) = 338 - x lo g_{10} (10)

2 \cdot lo g_{2} (x) = 4

lo g_{2} (lo g_{3} (lo g_{3} (b))) = 0

ln (x + 4) = - 2

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