What is factor-3ln|x+3|+3ln|x+1| ?

The solution to factor-3ln|x+3|+3ln|x+1| is ln((|x+1|^3)/(|x+3|^3))

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factor-3ln|x+3|+3ln|x+1|

factor

- 3 ln ∣ x + 3 ∣ + 3 ln ∣ x + 1 ∣

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

Solution steps

- 3 ln ∣ x + 3 ∣ + 3 ln ∣ x + 1 ∣

Apply log rule:

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

3 ln ∣ x + 3 ∣ = ln (∣ x + 3 ∣^{3})

= - ln (∣ x + 3 ∣^{3}) + 3 ln ∣ x + 1 ∣

Apply log rule:

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

3 ln ∣ x + 1 ∣ = ln (∣ x + 1 ∣^{3})

= - ln (∣ x + 3 ∣^{3}) + ln (∣ x + 1 ∣^{3})

Apply log rule:

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

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

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

\frac{1}{2} \cdot ln (\frac{0.697}{0.1828})

3 lo g_{10} (x) - 2 lo g_{10} (x + 8)

ln (\frac{1}{35.2 - 1})

- ln (u) + \frac{1}{2} \cdot ln (u - 1) + \frac{1}{2} \cdot ln (u - 1)

ln (\frac{1}{x ^{6}})

What is factor-3ln|x+3|+3ln|x+1| ?
The solution to factor-3ln|x+3|+3ln|x+1| is ln((|x+1|^3)/(|x+3|^3))