What is rationalize ln|x+2|-1/2 ln|y^2-1| ?

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

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rationalize ln|x+2|-1/2 ln|y^2-1|

rationalize

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Simplify

\frac{∣ x + 2 ∣}{∣ y ^{2} - 1 ∣ ^{\frac{1}{2}}} : \frac{∣ x + 2 ∣ ∣ y ^{2} - 1 ∣}{∣ y ^{2} - 1 ∣}

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

lo g_{10} (\frac{x ^{3} y}{z})

0.0296 \cdot 46 e^{(\frac{l n ( 263 ) - l n ( 46 )}{30}) 3}

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

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

ln (y) - ln (11 - y)

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