What is 1/2 ln|x-3|-1/(x-3)+1/2 ln|x+3|+C ?

The solution to 1/2 ln|x-3|-1/(x-3)+1/2 ln|x+3|+C is ln(|x-3|^{1/2}|x+3|^{1/2})-1/(x-3)+C

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simplify 1/2 ln|x-3|-1/(x-3)+1/2 ln|x+3|+C

simplify

\frac{1}{2} ln ∣ x - 3 ∣ - \frac{1}{x - 3} + \frac{1}{2} ln ∣ x + 3 ∣ + C

ln (∣ x - 3 ∣^{\frac{1}{2}} ∣ x + 3 ∣^{\frac{1}{2}}) - \frac{1}{x - 3} + C

Solution steps

\frac{1}{2} ln ∣ x - 3 ∣ - \frac{1}{x - 3} + \frac{1}{2} ln ∣ x + 3 ∣ + C

Apply log rule:

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

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

= ln (∣ x - 3 ∣^{\frac{1}{2}}) - \frac{1}{x - 3} + \frac{1}{2} ln ∣ x + 3 ∣ + C

Apply log rule:

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

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

= ln (∣ x - 3 ∣^{\frac{1}{2}}) - \frac{1}{x - 3} + ln (∣ x + 3 ∣^{\frac{1}{2}}) + C

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

ln (∣ x - 3 ∣^{\frac{1}{2}}) + ln (∣ x + 3 ∣^{\frac{1}{2}}) = ln (∣ x - 3 ∣^{\frac{1}{2}} ∣ x + 3 ∣^{\frac{1}{2}})

= ln (∣ x - 3 ∣^{\frac{1}{2}} ∣ x + 3 ∣^{\frac{1}{2}}) - \frac{1}{x - 3} + C

ln (5) + 3 ln (4) - ln (8)

- lo g_{10} (8.41 \cdot 1 0^{- 4})

ln (\frac{p}{x})

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

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

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