What is 2[3ln(x)-ln(x+1)-ln(x-1)] ?

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

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simplify 2[3ln(x)-ln(x+1)-ln(x-1)]

simplify

2 [3 ln (x) - ln (x + 1) - ln (x - 1)]

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

Solution steps

2 [3 ln (x) - ln (x + 1) - ln (x - 1)]

Apply log rule:

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

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

= 2 [ln (x^{3}) - ln (x + 1) - ln (x - 1)]

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply the fraction rule:

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

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

Remove parentheses:

(a) = a

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

- lo g_{b} (\frac{1}{2})

ln (x^{2} y^{5})

lo g_{x} (x x)

ln (3.7) - ln (1.9)

lo g_{10} (7 x^{3})

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