What is 1/5 ln|x-3|-1/5 ln|x+2| ?

The solution to 1/5 ln|x-3|-1/5 ln|x+2| is ln(\sqrt[5]{(|x-3|)/(|x+2|)})

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simplify 1/5 ln|x-3|-1/5 ln|x+2|

simplify

\frac{1}{5} ln ∣ x - 3 ∣ - \frac{1}{5} ln ∣ x + 2 ∣

Solution steps

\frac{1}{5} ln ∣ x - 3 ∣ - \frac{1}{5} ln ∣ x + 2 ∣

Apply log rule:

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

\frac{1}{5} ln ∣ x - 3 ∣ = ln (∣ x - 3 ∣^{\frac{1}{5}})

= ln (∣ x - 3 ∣^{\frac{1}{5}}) - \frac{1}{5} ln ∣ x + 2 ∣

Apply log rule:

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

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

= ln (∣ x - 3 ∣^{\frac{1}{5}}) - ln (∣ x + 2 ∣^{\frac{1}{5}})

Apply log rule:

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

ln (∣ x - 3 ∣^{\frac{1}{5}}) - ln (∣ x + 2 ∣^{\frac{1}{5}}) = ln (\frac{∣ x - 3 ∣ ^{\frac{1}{5}}}{∣ x + 2 ∣ ^{\frac{1}{5}}})

= ln (\frac{∣ x - 3 ∣ ^{\frac{1}{5}}}{∣ x + 2 ∣ ^{\frac{1}{5}}})

Combine same powers :

lo g_{e} (\frac{2}{3} w)

2 ln ∣ x + 2 ∣ - ln ∣ x - 4 ∣ + C

ln (θ^{y} (1 - θ^{n - y}))

lo g_{10} (5 x^{4} \cdot (1 - x)^{12 - 4})

What is 1/5 ln|x-3|-1/5 ln|x+2| ?
The solution to 1/5 ln|x-3|-1/5 ln|x+2| is ln(\sqrt[5]{(|x-3|)/(|x+2|)})