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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

2 ln ∣ x ∣ = ln (∣ x ∣^{2})

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

Apply log rule:

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

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

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

∣ x ∣^{2} = x^{2}

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

Apply log rule:

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

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

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

Multiply fractions:

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

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

22 (- \frac{1}{4} ln ∣ z + 2 ∣ + \frac{1}{4} ln ∣ z - 2 ∣)

- lo g_{10} (2.35 \cdot 1 0^{- 3})

\frac{19}{13} + \frac{4}{5} ln (\frac{5}{4}) + \frac{3}{8} ln (2)

- lo g_{10} (2.2 \cdot 1 0^{- 8})

ln (4 e^{17})

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