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

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

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

simplify

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

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

Solution steps

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

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

lo g_{e} (\frac{8}{3})

7.52 + lo g_{10} (\frac{0.02}{0.48})

lo g_{5} (\frac{a ^{5}}{b ^{8}})

lo g_{5} (\frac{x ^{2} ( 1 - 5 x ) ^{\frac{3}{2}}}{x ^{2} - x})

ln (\frac{21}{10})

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