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

The solution to 1/3 ln|x-1|-1/3 ln(x^2+x+1) is ln(\sqrt[3]{(|x-1|)/(x^2+x+1)})

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

simplify

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

Solution steps

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

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

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

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

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

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

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

Apply log rule:

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

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

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

Combine same powers :

4.74 - lo g_{10} (\frac{1}{1.8597})

ln (2 e^{- 1})

ln (e x p x)

10 lo g_{10} (9.64) - 10 lo g_{10} (0.37)

\frac{1}{2} ln (\frac{7 + 7}{3})

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