How do you condense $ln (x + 1) + ln (x - 1) - 3 ln (x)$ $ln(x+1)+ln(x-1)-3ln(x)$ ?

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Answer 1 · 2016-04-17T11:57:10

$ln (x^{- 1} - x^{- 3}) = ln (\frac{1}{x} - \frac{1}{x^{3}})$ $ln(x^-1-x^-3)=ln(1/x-1/x^3)$

Condense the last $log$ $log$ by recognising that a coefficient outside of the log is the same as an exponent inside, or

$3 ln x = {ln x}^{3}$ $3lnx=lnx^3$ .

Using the law that $log a + log b = log a b$ $loga+logb=logab$ , and $log a - log b = log (\frac{a}{b})$ $loga-logb=log(a/b)$ , then,

$ln (x + 1) + ln (x - 1) = ln [(x + 1) (x - 1)]$ $ln(x+1)+ln(x-1)=ln[(x+1)(x-1)]$
$= ln (x^{2} - 1)$ $=ln(x^2-1)$

and

$ln (x^{2} - 1) - {ln x}^{3} = ln (\frac{x^{2} - 1}{x^{3}})$ $ln(x^2-1)-lnx^3=ln((x^2-1)/x^3)$

which you could also condense, using laws of exponents, into

$ln (\frac{x^{2} - 1}{x^{3}}) = ln (x^{- 1} - x^{- 3})$ $ln((x^2-1)/x^3)=ln(x^-1-x^-3)$
$= ln (\frac{1}{x} - \frac{1}{x^{3}})$ $=ln(1/x-1/x^3)$

How do you condense ln(x+1)+ln(x-1)-3ln(x) ln(x+1)+ln(x−1)−3ln(x)ln(x+1)+ln(x-1)-3ln(x) ?