What is 1/2 (ln|x+1|-ln|x-1|) ?

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

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simplify 1/2 (ln|x+1|-ln|x-1|)

simplify

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

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

Solution steps

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

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

ln (2 a x)

lo g_{10} (3 x y)

ln (\frac{386}{530})

9 ln (a) - 7 ln (b)

ln (5) + ln (- 1)

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