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} (ab)

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

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

Multiply fractions:

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

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

Remove parentheses:

(a) = a

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

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

ln (\frac{0.8}{0. 8 ^{2}})

ln ∣ x + 4 ∣ - ln ∣ x + 2 ∣

lo g_{1.8} (\frac{6.69}{3.12}) + 53.29 - 54.01

10 lo g_{10} (\frac{2 \cdot 1 0 ^{- 6}}{1 0 ^{- 12}})

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))