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

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

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

simplify

ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1

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

Solution steps

ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Simplify

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

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

ln (\frac{76}{100})

ln (\frac{t y}{t x})

lo g_{10} (316) + 3 lo g_{10} (8 (5)) - 2.92

- lo g_{10} (6 \cdot 1 0^{- 8})

(\frac{( cos ( x ) )}{( 2 sin ^{2} ( x ) )}) + (\frac{1}{2}) ln (\frac{x}{2})

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