What is log_{10}(x^3+3x^2-4x)-log_{10}(x^2-2x+1) ?

The solution to log_{10}(x^3+3x^2-4x)-log_{10}(x^2-2x+1) is log_{10}((x(x+4))/(x-1))

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simplify log_{10}(x^3+3x^2-4x)-log_{10}(x^2-2x+1)

simplify

lo g_{10} (x^{3} + 3 x^{2} - 4 x) - lo g_{10} (x^{2} - 2 x + 1)

lo g_{10} (\frac{x ( x + 4 )}{x - 1})

Solution steps

lo g_{10} (x^{3} + 3 x^{2} - 4 x) - lo g_{10} (x^{2} - 2 x + 1)

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{10} (x^{3} + 3 x^{2} - 4 x) - lo g_{10} (x^{2} - 2 x + 1) = lo g_{10} (\frac{x ^{3} + 3 x ^{2} - 4 x}{x ^{2} - 2 x + 1})

= lo g_{10} (\frac{x ^{3} + 3 x ^{2} - 4 x}{x ^{2} - 2 x + 1})

Simplify

\frac{x ^{3} + 3 x ^{2} - 4 x}{x ^{2} - 2 x + 1} : \frac{x ( x + 4 )}{x - 1}

= lo g_{10} (\frac{x ( x + 4 )}{x - 1})

1.344 \cdot ln (\frac{495}{639})

\frac{2}{7} lo g_{10} (a) - \frac{1}{4} lo g_{10} (b)

- lo g_{10} (2.68 \cdot 1 0^{- 3})

ln (\frac{4 x ^{2} - x - 1}{3 x ^{2} + 5})

lo g_{10} (X^{6} Y)

What is log_{10}(x^3+3x^2-4x)-log_{10}(x^2-2x+1) ?
The solution to log_{10}(x^3+3x^2-4x)-log_{10}(x^2-2x+1) is log_{10}((x(x+4))/(x-1))