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 )})

f (x) = ln (6 (ln (x))^{8})

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

ln (\frac{729}{625})

f (x) = lo g_{10} (x^{2} - 5 x + 6) - lo g_{10} (x - 3)

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)))