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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

ln ∣ z + 4 ∣ + ln ∣ x ∣ = ln (∣ x ∣ ∣ z + 4 ∣)

= ln (∣ z + 4 ∣ ∣ x ∣) + 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 (∣ x ∣ ∣ 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 (∣ x ∣ ∣ 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 (∣ x ∣ ∣ 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}} ∣ x ∣ ∣ z + 4 ∣)

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

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

lo g_{\frac{4}{3}} (- 3 x) + 6

- lo g_{10} (2.07 \cdot 1 0^{- 6})

\frac{7}{2} lo g_{10} (a) + 4 lo g_{10} (b)

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