What is 1/3 [2ln(x+5)-ln(x)-ln(x^2-4)] ?

The solution to 1/3 [2ln(x+5)-ln(x)-ln(x^2-4)] is 1/3 ln(((x+5)^2)/(x(x^2-4)))

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simplify 1/3 [2ln(x+5)-ln(x)-ln(x^2-4)]

simplify

\frac{1}{3} [2 ln (x + 5) - ln (x) - ln (x^{2} - 4)]

\frac{1}{3} ln (\frac{( x + 5 ) ^{2}}{x ( x ^{2} - 4 )})

Solution steps

\frac{1}{3} [2 ln (x + 5) - ln (x) - ln (x^{2} - 4)]

Apply log rule:

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

2 ln (x + 5) = ln ((x + 5)^{2})

= \frac{1}{3} [ln ((x + 5)^{2}) - ln (x) - ln (x^{2} - 4)]

Apply log rule:

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

ln ((x + 5)^{2}) - ln (x) = ln (\frac{( x + 5 ) ^{2}}{x})

= \frac{1}{3} [ln (\frac{( x + 5 ) ^{2}}{x}) - ln (x^{2} - 4)]

Apply log rule:

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

ln (\frac{( x + 5 ) ^{2}}{x}) - ln (x^{2} - 4) = ln (\frac{\frac{( x + 5 ) ^{2}}{x}}{x ^{2} - 4})

= \frac{1}{3} [ln (\frac{\frac{( x + 5 ) ^{2}}{x}}{x ^{2} - 4})]

Apply the fraction rule:

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

= \frac{1}{3} [ln (\frac{( x + 5 ) ^{2}}{x ( x ^{2} - 4 )})]

Remove parentheses:

(a) = a

= \frac{1}{3} ln (\frac{( x + 5 ) ^{2}}{x ( x ^{2} - 4 )})

lo g_{3} (7) + lo g_{3} (5)

\frac{1}{2} lo g_{6} (9) + lo g_{6} (5)

2 lo g_{b} (y) + 6 lo g_{b} (z)

ln^{2} (x) - ln (x)

ln (x) - 4 [ln (x + 2) + ln (x - 2)]

What is 1/3 [2ln(x+5)-ln(x)-ln(x^2-4)] ?
The solution to 1/3 [2ln(x+5)-ln(x)-ln(x^2-4)] is 1/3 ln(((x+5)^2)/(x(x^2-4)))