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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Simplify

\frac{\frac{∣ v - 1 ∣}{∣ v + 1 ∣}}{∣ v ^{2} - 1 ∣ ^{\frac{1}{2}}} : \frac{∣ v - 1 ∣ ∣ v ^{2} - 1 ∣}{∣ v + 1 ∣ ∣ v ^{2} - 1 ∣}

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

ln (23)

ln (a \cdot e)

lo g_{4} (\frac{64 x ^{2}}{y})

16.7 + 10 lo g_{10} (\frac{10400}{120})

ln (\frac{4 x ^{2}}{( x + 3 ) ^{3}})

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