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

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

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

simplify

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

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

Solution steps

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

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 ∣ + C

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 ∣ + C

∣ x + 1 ∣^{2} = (x + 1)^{2}

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

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 ∣) + C

Multiply fractions:

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

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

lo g_{3} (\frac{81}{y})

2 ln (3) - 2 ln (2)

lo g_{2} (\frac{8 X ^{3}}{2 Y})

lo g_{10} (a) + 3 lo g_{10} (2 b)

ln (\frac{4}{e})

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