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

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

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

simplify

\frac{1}{2} ln ∣ y - 1 ∣ - \frac{1}{2} ln ∣ y + 1 ∣

ln (\frac{∣ y - 1 ∣}{∣ y + 1 ∣})

Solution steps

\frac{1}{2} ln ∣ y - 1 ∣ - \frac{1}{2} ln ∣ y + 1 ∣

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{2} ln ∣ y - 1 ∣ = ln (∣ y - 1 ∣^{\frac{1}{2}})

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

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{2} ln ∣ y + 1 ∣ = ln (∣ y + 1 ∣^{\frac{1}{2}})

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

Apply log rule:

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

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

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

Combine same powers :

= ln (\frac{∣ y - 1 ∣}{∣ y + 1 ∣})

lo g_{10} (2) - lo g_{10} (20.2)

- 50 ln (\frac{2}{3})

- lo g_{10} (1.17 \cdot 1 0^{- 6})

ln (\frac{20000}{477})

ln (θ \cdot e^{- θ x})

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