What is 3(-1/6 ln|x+2|+1/6 ln|x-4|)+C ?

The solution to 3(-1/6 ln|x+2|+1/6 ln|x-4|)+C is 3ln(\sqrt[6]{(|x-4|)/(|x+2|)})+C

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simplify 3(-1/6 ln|x+2|+1/6 ln|x-4|)+C

simplify

3 (- \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 4 ∣) + C

Solution steps

3 (- \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 4 ∣) + C

Apply log rule:

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

\frac{1}{6} ln ∣ x + 2 ∣ = ln (∣ x + 2 ∣^{\frac{1}{6}})

= 3 (\frac{1}{6} ln ∣ x - 4 ∣ - ln (∣ x + 2 ∣^{\frac{1}{6}})) + C

Apply log rule:

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

\frac{1}{6} ln ∣ x - 4 ∣ = ln (∣ x - 4 ∣^{\frac{1}{6}})

= 3 (ln (∣ x - 4 ∣^{\frac{1}{6}}) - ln (∣ x + 2 ∣^{\frac{1}{6}})) + C

Apply log rule:

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

ln (∣ x - 4 ∣^{\frac{1}{6}}) - ln (∣ x + 2 ∣^{\frac{1}{6}}) = ln (\frac{∣ x - 4 ∣ ^{\frac{1}{6}}}{∣ x + 2 ∣ ^{\frac{1}{6}}})

= 3 (ln (\frac{∣ x - 4 ∣ ^{\frac{1}{6}}}{∣ x + 2 ∣ ^{\frac{1}{6}}})) + C

Combine same powers :

Remove parentheses:

(a) = a

lo g_{3} ((2 x)^{2})

ln (\frac{1}{a b ^{3}})

ln ((\frac{2}{a ^{3}})^{4})

lo g_{10} (\frac{a ^{4}}{b ^{2} c ^{3}})

ln (θ (θ + 1) x^{θ - 1} (1 - x))

What is 3(-1/6 ln|x+2|+1/6 ln|x-4|)+C ?
The solution to 3(-1/6 ln|x+2|+1/6 ln|x-4|)+C is 3ln(\sqrt[6]{(|x-4|)/(|x+2|)})+C