What is ln|1-x|+1/2 ln|x^2+16|+C ?

The solution to ln|1-x|+1/2 ln|x^2+16|+C is ln(|1-x||x^2+16|^{1/2})+C

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simplify ln|1-x|+1/2 ln|x^2+16|+C

simplify

ln ∣ 1 - x ∣ + \frac{1}{2} ln x^{2} + 16 + C

ln (∣ 1 - x ∣ x^{2} + 16^{\frac{1}{2}}) + C

Solution steps

ln ∣ 1 - x ∣ + \frac{1}{2} ln x^{2} + 16 + C

Apply log rule:

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

\frac{1}{2} ln x^{2} + 16 = ln (x^{2} + 16^{\frac{1}{2}})

= ln ∣ - x + 1 ∣ + ln (x^{2} + 16^{\frac{1}{2}}) + C

Apply log rule:

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

ln ∣ 1 - x ∣ + ln (x^{2} + 16^{\frac{1}{2}}) = ln (∣ - x + 1 ∣ x^{2} + 16^{\frac{1}{2}})

= ln (∣ 1 - x ∣ x^{2} + 16^{\frac{1}{2}}) + C

ln (\frac{7 ^{\frac{3}{4}}}{7})

ln (\frac{s - 2}{s - 1})

x (ln (x^{3}) + ln (e x))

4 lo g_{10} (x) + 8 lo g_{10} (y)

7 + lo g_{10} (\frac{0.35}{1.7})

What is ln|1-x|+1/2 ln|x^2+16|+C ?
The solution to ln|1-x|+1/2 ln|x^2+16|+C is ln(|1-x||x^2+16|^{1/2})+C