What is 6[ln(x)-ln(x+1)-ln(x-1)] ?

The solution to 6[ln(x)-ln(x+1)-ln(x-1)] is 6ln(x/((x+1)(x-1)))

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simplify 6[ln(x)-ln(x+1)-ln(x-1)]

simplify

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

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

Solution steps

6 [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})

= 6 [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})

= 6 [ln (\frac{\frac{x}{x + 1}}{x - 1})]

Apply the fraction rule:

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

= 6 [ln (\frac{x}{( x + 1 ) ( x - 1 )})]

Remove parentheses:

(a) = a

= 6 ln (\frac{x}{( x + 1 ) ( x - 1 )})

100 \cdot ln (209) - ln (193)

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- 2 ln (\frac{3}{2})

ln (\frac{( 40 - 50 )}{( 40 - 83.3 )})

ln (23 x)

What is 6[ln(x)-ln(x+1)-ln(x-1)] ?
The solution to 6[ln(x)-ln(x+1)-ln(x-1)] is 6ln(x/((x+1)(x-1)))