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 ∣})]

Remove parentheses:

(a) = a

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

lo g_{2} (10) + lo g_{2} (20)

lo g_{c} (\frac{a ^{2} c}{b ^{3}})

2 lo g_{2} (x) - lo g_{2} (x + 3)

\frac{3 ln ( \frac{7}{4} )}{ln ( 7 ) - 2 ln ( 2 )} - 2

lo g_{5} (14 x + 15) - lo g_{5} (\frac{1}{14 x - 15})

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|))