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

The solution to log_{e}(x^3-1)-log_{e}(x^2+x+1) is ln(x-1)

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

simplify

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

ln (x - 1)

Solution steps

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

Apply log rule:

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

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

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

Simplify

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

= lo g_{e} (x - 1)

Apply log rule:

lo g_{e} (x) = ln (x)

= ln (x - 1)

4 (lo g_{a} (w) - 4 lo g_{a} (z)) + 2 lo g_{a} (y)

3 ln_{7} (4) + ln_{7} (6)

lo g_{1.8} (\frac{8.13}{3.12}) + 52.11 - 54.01

lo g_{10} (4 p^{3})

- lo g_{10} (5.34 \cdot 1 0^{- 3})

What is log_{e}(x^3-1)-log_{e}(x^2+x+1) ?
The solution to log_{e}(x^3-1)-log_{e}(x^2+x+1) is ln(x-1)