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

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Answer 1 · 2016-04-17T11:57:10

$ln(x^-1-x^-3)=ln(1/x-1/x^3)$

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

$3lnx=lnx^3$ .

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

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

and

$ln(x^2-1)-lnx^3=ln((x^2-1)/x^3)$

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

$ln((x^2-1)/x^3)=ln(x^-1-x^-3)$
$=ln(1/x-1/x^3)$

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