What is 1/5 ln|x+1|+4/5 ln|x-4| ?

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

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simplify 1/5 ln|x+1|+4/5 ln|x-4|

simplify

\frac{1}{5} ln ∣ x + 1 ∣ + \frac{4}{5} ln ∣ x - 4 ∣

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

Solution steps

\frac{1}{5} ln ∣ x + 1 ∣ + \frac{4}{5} ln ∣ x - 4 ∣

Apply log rule:

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

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

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

Apply log rule:

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

\frac{4}{5} ln ∣ x - 4 ∣ = ln (∣ x - 4 ∣^{\frac{4}{5}})

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

Apply log rule:

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

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

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

5 lo g_{b} (t) + 2 lo g_{b} (w)

ln (a e^{x}) + ln (b e^{- x})

ln (x d x)

ln (\frac{x ^{6}}{y ^{2}})

ln (c) + ln (a)

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