What is (-1)/4 (ln| x/2+1|-ln| x/2-1|) ?

The solution to (-1)/4 (ln| x/2+1|-ln| x/2-1|) is -1/4 ln((|x+2|)/(|x-2|))

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simplify (-1)/4 (ln| x/2+1|-ln| x/2-1|)

simplify

\frac{- 1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

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

Solution steps

\frac{- 1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

Apply log rule:

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

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

= \frac{- 1}{4} ln (\frac{\frac{x}{2} + 1}{\frac{x}{2} - 1})

Simplify

\frac{\frac{x}{2} + 1}{\frac{x}{2} - 1} : \frac{∣ x + 2 ∣}{∣ x - 2 ∣}

= \frac{- 1}{4} ln (\frac{∣ x + 2 ∣}{∣ x - 2 ∣})

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{4} ln (\frac{∣ x + 2 ∣}{∣ x - 2 ∣})

- lo g_{10} (1.8 \cdot 1 0^{- 3})

ln (x - 155) - ln (77 - 155)

2 lo g_{10} (a) + 3 lo g_{10} (b) - 5 lo g_{10} (c)

ln (60 e^{- 8.064})

ln (\frac{7.28 \cdot 1 0 ^{- 4}}{1.93 \cdot 1 0 ^{- 7}})

What is (-1)/4 (ln| x/2+1|-ln| x/2-1|) ?
The solution to (-1)/4 (ln| x/2+1|-ln| x/2-1|) is -1/4 ln((|x+2|)/(|x-2|))