What is 3ln|x|-2ln|x+1| ?

The solution to 3ln|x|-2ln|x+1| is ln((|x|^3)/((x+1)^2))

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simplify 3ln|x|-2ln|x+1|

simplify

3 ln ∣ x ∣ - 2 ln ∣ x + 1 ∣

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

Solution steps

3 ln ∣ x ∣ - 2 ln ∣ x + 1 ∣

Apply log rule:

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

3 ln ∣ x ∣ = ln (∣ x ∣^{3})

= ln (∣ x ∣^{3}) - 2 ln ∣ x + 1 ∣

Apply log rule:

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

2 ln ∣ x + 1 ∣ = ln (∣ x + 1 ∣^{2})

= ln (∣ x ∣^{3}) - ln (∣ x + 1 ∣^{2})

Apply log rule:

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

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

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

∣ x + 1 ∣^{2} = (x + 1)^{2}

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

ln (a^{4} b^{7})

\frac{1}{2} lo g_{7} (x) + \frac{1}{3} lo g_{7} (y)

2 ln (5) + 2 ln (x) - ln (y)

ln (\frac{x ^{3} ln ( 2 x )}{e ^{x} sin ( x )})

ln (- i e^{3})

What is 3ln|x|-2ln|x+1| ?
The solution to 3ln|x|-2ln|x+1| is ln((|x|^3)/((x+1)^2))