What is (ln(x^3-1)-ln(x^2-x+1))/(ln(x+1)) ?

The solution to (ln(x^3-1)-ln(x^2-x+1))/(ln(x+1)) is (ln(\frac{x^3-1)/(x^2-x+1))}{ln(x+1)}

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simplify (ln(x^3-1)-ln(x^2-x+1))/(ln(x+1))

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

ln (\frac{m}{8} (p - 9))

4 - \frac{3}{2} + ln \frac{3}{2} - ln ∣ 4 ∣

lo g_{b} (y^{2} x)

ln (\frac{1}{x - 1})

ln (x^{2} - 64) - ln (x + 8)

What is (ln(x^3-1)-ln(x^2-x+1))/(ln(x+1)) ?
The solution to (ln(x^3-1)-ln(x^2-x+1))/(ln(x+1)) is (ln(\frac{x^3-1)/(x^2-x+1))}{ln(x+1)}