What is 4(1/9 ln|x-2|-1/9 ln|x+7|+C) ?

The solution to 4(1/9 ln|x-2|-1/9 ln|x+7|+C) is 4(ln(\sqrt[9]{(|x-2|)/(|x+7|)})+C)

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simplify 4(1/9 ln|x-2|-1/9 ln|x+7|+C)

simplify

4 (\frac{1}{9} ln ∣ x - 2 ∣ - \frac{1}{9} ln ∣ x + 7 ∣ + C)

Solution steps

4 (\frac{1}{9} ln ∣ x - 2 ∣ - \frac{1}{9} ln ∣ x + 7 ∣ + C)

Apply log rule:

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

\frac{1}{9} ln ∣ x - 2 ∣ = ln (∣ x - 2 ∣^{\frac{1}{9}})

= 4 (ln (∣ x - 2 ∣^{\frac{1}{9}}) - \frac{1}{9} ln ∣ x + 7 ∣ + C)

Apply log rule:

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

\frac{1}{9} ln ∣ x + 7 ∣ = ln (∣ x + 7 ∣^{\frac{1}{9}})

= 4 (ln (∣ x - 2 ∣^{\frac{1}{9}}) - ln (∣ x + 7 ∣^{\frac{1}{9}}) + C)

Apply log rule:

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

ln (∣ x - 2 ∣^{\frac{1}{9}}) - ln (∣ x + 7 ∣^{\frac{1}{9}}) = ln (\frac{∣ x - 2 ∣ ^{\frac{1}{9}}}{∣ x + 7 ∣ ^{\frac{1}{9}}})

= 4 (ln (\frac{∣ x - 2 ∣ ^{\frac{1}{9}}}{∣ x + 7 ∣ ^{\frac{1}{9}}}) + C)

Combine same powers :

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

lo g_{10} (2) - lo g_{10} (1.035)

\frac{1}{2} lo g_{c} (x) + 3 lo g_{c} (y) - 5 lo g_{c} (x)

- 2 - lo g_{4} (z) - lo g_{4} (x^{3} - 27)

lo g_{10} (- 3 i)

What is 4(1/9 ln|x-2|-1/9 ln|x+7|+C) ?
The solution to 4(1/9 ln|x-2|-1/9 ln|x+7|+C) is 4(ln(\sqrt[9]{(|x-2|)/(|x+7|)})+C)