What is 1/5 ln((x-1)/(x^2-x+1)) ?

The solution to 1/5 ln((x-1)/(x^2-x+1)) is 1/5 (ln(x-1)-ln(x^2-x+1))

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simplify 1/5 ln((x-1)/(x^2-x+1))

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

lo g_{10} (\frac{2 x ( 1 - 5 x ) ^{\frac{3}{2}}}{( x + 4 ) ^{3}})

ln (\frac{0.075}{0.06})

ln (\frac{2}{e ^{(- 1)} - e})

19 x + ln (19) + ln (x)

4 ln (4 x + 8) + 3 ln (2 x + 7)

What is 1/5 ln((x-1)/(x^2-x+1)) ?
The solution to 1/5 ln((x-1)/(x^2-x+1)) is 1/5 (ln(x-1)-ln(x^2-x+1))