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

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

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

simplify

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

lo g_{10} (\frac{y - 2}{y + 1})

Solution steps

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

Apply log rule:

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

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

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

Simplify

\frac{y ^{2} - 3 y + 2}{y ^{2} - 1} : \frac{y - 2}{y + 1}

= lo g_{10} (\frac{y - 2}{y + 1})

ln (\frac{3859}{35109})

lo g_{2} (\frac{64}{250}) + 6 lo g_{4} (5)

lo g_{2} (x y)

lo g_{10} (\frac{x ^{2}}{5})

- lo g_{10} (1 \cdot 1 0^{- 9})

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