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^2)/(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 ^{2}}{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 )}

Subtract the numbers:

3 - 1 = 2

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

lo g_{3} (4) - lo g_{3} (7)

\frac{1}{2} lo g_{3} (5 x)

\frac{1}{3} (3 lo g_{7} (a) + 2 lo g_{7} (b))

\frac{2}{3} ln (8) - 2 ln (4) + ln (24)

[\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]

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^2)/(x^2-x+1))}{ln(x+1)}