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

The solution to ln|z+4|+6(1/5 ln|z-1|-1/5 ln|z+4|) is ln(((|z-1|)/(|z+4|))^{6/5}|z+4|)

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

simplify

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

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

Solution steps

ln ∣ z + 4 ∣ + 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}})

= ln ∣ z + 4 ∣ + 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}})

= ln ∣ z + 4 ∣ + 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}}})

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

Combine same powers :

Apply log rule:

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

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

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

3 ln (3 x)

ln (3 (10 x^{2})^{x})

ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣

ln (17) - ln (16)

2 ln (3) + 4 ln (2) - ln (36)

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