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

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

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

simplify

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

ln (\frac{∣ x - 1 ∣}{∣ x + 1 ∣}) + C

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Combine same powers :

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

lo g_{5} (\frac{125}{x - 1})

ln (x (x - 3))

- lo g_{10} (7.7 \cdot 1 0^{- 3})

f (x) = 2 ln (x^{3}) - 3 ln (x)

ln (x e^{- 2 x})

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