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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

2 ln ∣ x - 1 ∣ = ln (∣ x - 1 ∣^{2})

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

Apply log rule:

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

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

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

∣ x - 1 ∣^{2} = (x - 1)^{2}

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

\frac{π}{2} ln (3)

lo g_{10} (x) - 4 lo g_{10} (y)

ln (e^{x^{2}} cos (5 x - 7))

lo g_{3} (8) + lo g_{3} (x) - 2 lo g_{3} (x + 4)

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

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