What is 1/4 [3ln(x+9)-ln(x)-ln(x^2-36)] ?

The solution to 1/4 [3ln(x+9)-ln(x)-ln(x^2-36)] is 1/4 ln(((x+9)^3)/(x(x^2-36)))

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simplify 1/4 [3ln(x+9)-ln(x)-ln(x^2-36)]

simplify

\frac{1}{4} [3 ln (x + 9) - ln (x) - ln (x^{2} - 36)]

\frac{1}{4} ln (\frac{( x + 9 ) ^{3}}{x ( x ^{2} - 36 )})

Solution steps

\frac{1}{4} [3 ln (x + 9) - ln (x) - ln (x^{2} - 36)]

Apply log rule:

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

3 ln (x + 9) = ln ((x + 9)^{3})

= \frac{1}{4} [ln ((x + 9)^{3}) - ln (x) - ln (x^{2} - 36)]

Apply log rule:

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

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

= \frac{1}{4} [ln (\frac{( x + 9 ) ^{3}}{x}) - ln (x^{2} - 36)]

Apply log rule:

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

ln (\frac{( x + 9 ) ^{3}}{x}) - ln (x^{2} - 36) = ln (\frac{\frac{( x + 9 ) ^{3}}{x}}{x ^{2} - 36})

= \frac{1}{4} [ln (\frac{\frac{( x + 9 ) ^{3}}{x}}{x ^{2} - 36})]

Apply the fraction rule:

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

= \frac{1}{4} [ln (\frac{( x + 9 ) ^{3}}{x ( x ^{2} - 36 )})]

Remove parentheses:

(a) = a

= \frac{1}{4} ln (\frac{( x + 9 ) ^{3}}{x ( x ^{2} - 36 )})

lo g_{10} (\frac{x ^{2}}{2 z})

lo g_{b} (u^{8} v^{5})

lo g_{7} (\frac{7 x ^{5}}{y})

lo g_{3} (3 (5 x))

2 lo g_{10} (5) + 3 lo g_{10} (2)

What is 1/4 [3ln(x+9)-ln(x)-ln(x^2-36)] ?
The solution to 1/4 [3ln(x+9)-ln(x)-ln(x^2-36)] is 1/4 ln(((x+9)^3)/(x(x^2-36)))