What is 6(1/5 ln|z-1|-1/5 ln|z+4|) ?

The solution to 6(1/5 ln|z-1|-1/5 ln|z+4|) is 6ln(\sqrt[5]{(|z-1|)/(|z+4|)})

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simplify 6(1/5 ln|z-1|-1/5 ln|z+4|)

simplify

6 (\frac{1}{5} ln ∣ z - 1 ∣ - \frac{1}{5} ln ∣ z + 4 ∣)

Solution steps

6 (\frac{1}{5} ln ∣ z - 1 ∣ - \frac{1}{5} ln ∣ z + 4 ∣)

Apply log rule:

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

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

= 6 (ln (∣ z - 1 ∣^{\frac{1}{5}}) - \frac{1}{5} ln ∣ z + 4 ∣)

Apply log rule:

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

\frac{1}{5} ln ∣ z + 4 ∣ = ln (∣ z + 4 ∣^{\frac{1}{5}})

= 6 (ln (∣ z - 1 ∣^{\frac{1}{5}}) - ln (∣ z + 4 ∣^{\frac{1}{5}}))

Apply log rule:

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

ln (∣ z - 1 ∣^{\frac{1}{5}}) - ln (∣ z + 4 ∣^{\frac{1}{5}}) = ln (\frac{∣ z - 1 ∣ ^{\frac{1}{5}}}{∣ z + 4 ∣ ^{\frac{1}{5}}})

= 6 ln (\frac{∣ z - 1 ∣ ^{\frac{1}{5}}}{∣ z + 4 ∣ ^{\frac{1}{5}}})

Combine same powers :

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

- ln (i - n) + ln (i)

3 ln (\frac{3}{2})

ln (t) + ln (t + 6) + ln (t + 4)

\frac{8 \cdot 12 5 ^{- x}}{3 ln ( 5 )} + C

What is 6(1/5 ln|z-1|-1/5 ln|z+4|) ?
The solution to 6(1/5 ln|z-1|-1/5 ln|z+4|) is 6ln(\sqrt[5]{(|z-1|)/(|z+4|)})