What is 1/3 (ln(x)+ln(x-6)-ln(x^3+9)) ?

The solution to 1/3 (ln(x)+ln(x-6)-ln(x^3+9)) is 1/3 ln((x(x-6))/(x^3+9))

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simplify 1/3 (ln(x)+ln(x-6)-ln(x^3+9))

simplify

\frac{1}{3} (ln (x) + ln (x - 6) - ln (x^{3} + 9))

\frac{1}{3} ln (\frac{x ( x - 6 )}{x ^{3} + 9})

Solution steps

\frac{1}{3} (ln (x) + ln (x - 6) - ln (x^{3} + 9))

Apply log rule:

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

ln (x) + ln (x - 6) = ln (x (x - 6))

= \frac{1}{3} (ln (x (x - 6)) - ln (x^{3} + 9))

Apply log rule:

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

ln (x (x - 6)) - ln (x^{3} + 9) = ln (\frac{x ( x - 6 )}{x ^{3} + 9})

= \frac{1}{3} ln (\frac{x ( x - 6 )}{x ^{3} + 9})

\frac{1}{2} lo g_{10} (x + 3) + \frac{1}{2} lo g_{10} (2 x - 1)

\frac{ln ( \frac{69}{97} )}{2 ln ( 2 )}

\frac{1}{5} lo g_{3} (10)

- \frac{1}{16} lo g_{e} (\frac{1}{16}) - \frac{15}{16} lo g_{e} (\frac{15}{16})

ln (2 \cdot e^{- 10 k})

What is 1/3 (ln(x)+ln(x-6)-ln(x^3+9)) ?
The solution to 1/3 (ln(x)+ln(x-6)-ln(x^3+9)) is 1/3 ln((x(x-6))/(x^3+9))