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

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

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

simplify

ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x - 2 ∣

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

Solution steps

ln ∣ x ∣ - 2 ln ∣ x + 1 ∣ + ln ∣ x - 2 ∣

Apply log rule:

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

2 ln ∣ x + 1 ∣ = ln (∣ x + 1 ∣^{2})

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

Apply log rule:

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

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

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

∣ x + 1 ∣^{2} = (x + 1)^{2}

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

Apply log rule:

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

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

ln (3) - ln (\frac{1}{3})

ln (x) - 2 [ln (x - 7) + ln (x + 7)]

4 lo g_{b} (x) + \frac{1}{2} lo g_{b} (y) - 3 lo g_{b} (2 z)

lo g_{a} (\frac{3 ^{4}}{4 ^{2} a ^{7}})

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