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

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

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

simplify

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

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

Solution steps

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

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

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

Apply the fraction rule:

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

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

Remove parentheses:

(a) = a

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

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

lo g_{3} (c) + lo g_{3} (c + 3)

lo g_{5} (\frac{a ^{2} b ^{2} x}{c ^{3}})

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

lo g_{5} (x) - lo g_{5} (x^{8})

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