What is 1/4 [3ln(x+4)-ln(x)-ln(x^2-25)] ?

The solution to 1/4 [3ln(x+4)-ln(x)-ln(x^2-25)] is 1/4 ln(((x+4)^3)/(x(x^2-25)))

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simplify 1/4 [3ln(x+4)-ln(x)-ln(x^2-25)]

simplify

\frac{1}{4} [3 ln (x + 4) - ln (x) - ln (x^{2} - 25)]

\frac{1}{4} ln (\frac{( x + 4 ) ^{3}}{x ( x ^{2} - 25 )})

Solution steps

\frac{1}{4} [3 ln (x + 4) - ln (x) - ln (x^{2} - 25)]

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} - 25)]

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} - 25)]

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} - 25) = ln (\frac{\frac{( x + 4 ) ^{3}}{x}}{x ^{2} - 25})

= \frac{1}{4} [ln (\frac{\frac{( x + 4 ) ^{3}}{x}}{x ^{2} - 25})]

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} - 25 )})]

Remove parentheses:

(a) = a

= \frac{1}{4} ln (\frac{( x + 4 ) ^{3}}{x ( x ^{2} - 25 )})

- ln (x) + ln (x + 1)

lo g_{2} (x^{2} y^{3}) - lo g_{2} (x) - lo g_{2} (y)

3 lo g_{10} (x \cdot x^{3})

\frac{1}{2} lo g_{10} (16) - \frac{1}{3} lo g_{10} (8) + 1

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

What is 1/4 [3ln(x+4)-ln(x)-ln(x^2-25)] ?
The solution to 1/4 [3ln(x+4)-ln(x)-ln(x^2-25)] is 1/4 ln(((x+4)^3)/(x(x^2-25)))