What is 1/4 [3ln(x+4)-ln(x)-ln(x^2-9)] ?

The solution to 1/4 [3ln(x+4)-ln(x)-ln(x^2-9)] is 1/4 ln(((x+4)^3)/(x(x^2-9)))

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simplify 1/4 [3ln(x+4)-ln(x)-ln(x^2-9)]

simplify

\frac{1}{4} [3 ln (x + 4) - ln (x) - ln (x^{2} - 9)]

\frac{1}{4} ln (\frac{( x + 4 ) ^{3}}{x ( x ^{2} - 9 )})

Solution steps

\frac{1}{4} [3 ln (x + 4) - ln (x) - ln (x^{2} - 9)]

Apply log rule:

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

3 ln (x + 4) = ln ((x + 4)^{3})

= \frac{1}{4} [ln ((x + 4)^{3}) - ln (x) - ln (x^{2} - 9)]

Apply log rule:

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

ln ((x + 4)^{3}) - ln (x) = ln (\frac{( x + 4 ) ^{3}}{x})

= \frac{1}{4} [ln (\frac{( x + 4 ) ^{3}}{x}) - ln (x^{2} - 9)]

Apply log rule:

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

ln (\frac{( x + 4 ) ^{3}}{x}) - ln (x^{2} - 9) = ln (\frac{\frac{( x + 4 ) ^{3}}{x}}{x ^{2} - 9})

= \frac{1}{4} [ln (\frac{\frac{( x + 4 ) ^{3}}{x}}{x ^{2} - 9})]

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{1}{4} [ln (\frac{( x + 4 ) ^{3}}{x ( x ^{2} - 9 )})]

Remove parentheses:

(a) = a

= \frac{1}{4} ln (\frac{( x + 4 ) ^{3}}{x ( x ^{2} - 9 )})

lo g_{5} (\frac{1}{625})

ln ((x + 3)^{2} (x - 1)^{3}) + \frac{4}{( x - 1 )} + c

- lo g_{10} (3.8 (1 0^{- 5}))

ln (\frac{10}{22})

lo g_{\frac{1}{7}} (3)

What is 1/4 [3ln(x+4)-ln(x)-ln(x^2-9)] ?
The solution to 1/4 [3ln(x+4)-ln(x)-ln(x^2-9)] is 1/4 ln(((x+4)^3)/(x(x^2-9)))