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

The solution to 1/2 ln|1^2+1|-1/2 ln|-2^2+1| is 1/2 ln(2/3)

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

simplify

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

\frac{1}{2} ln (\frac{2}{3})

Decimal Notation

- 0.20273 \dots

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

\frac{1 ^{2} + 1 ^{\frac{1}{2}}}{∣ - 2 ^{2} + 1 ∣ ^{\frac{1}{2}}} = \frac{2}{3}

= ln (\frac{2}{3})

Rewrite as

= ln ((\frac{2}{3})^{\frac{1}{2}})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= \frac{1}{2} ln (\frac{2}{3})

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What is 1/2 ln|1^2+1|-1/2 ln|-2^2+1| ?
The solution to 1/2 ln|1^2+1|-1/2 ln|-2^2+1| is 1/2 ln(2/3)