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} (ab)

ln (x) + ln (x + 1) = ln (x (x + 1))

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

Apply log rule:

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

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

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

Remove parentheses:

(a) = a

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

lo g_{3} (\frac{16}{z})

lo g_{10} (4 x^{2} y)

ln^{4} (a x)

lo g_{10} (\frac{y ^{3}}{2 x})

ln (\frac{1}{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))