What is simplify-1/2 ln|v+2|+1/2 ln|v| ?

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

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

simplify

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

ln (\frac{∣ v ∣}{∣ v + 2 ∣})

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Combine same powers :

= ln (\frac{∣ v ∣}{∣ v + 2 ∣})

ln (\frac{1}{a} e^{- \frac{y}{a}})

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