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

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

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

simplify

ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1

ln (\frac{∣ x + 2 ∣}{∣ x - 2 ∣})

Solution steps

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})

= 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 ∣}

= ln (\frac{∣ x + 2 ∣}{∣ x - 2 ∣})

lo g_{10} (12 x) - lo g_{10} (3 x)

ln (4 h - 8) - ln (4)

\frac{1}{2} ln (x) + 6 ln (y) - 5 ln (z)

- lo g_{10} (4.24 \cdot 1 0^{- 3})

lo g_{10} (y x^{9})

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