Simplify $ln(x/(x-1))+ ln((x+1)/x)-ln(x^2-1)$ ?

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Answer 1 · 2017-04-09T17:30:43

$-2log(x-1)$

Using the logarithmic properties

$log(a b)=loga + log b$
$loga^b = b log a$

we have

$log(x/(x-1))+log((x+1)/x)-log(x^2-1)=log(((x/(x-1))((x+1)/x))/(x^2-1))$

$=log(((x+1)/(x-1))/((x+1)(x-1)))=log(1/(x-1)^2)=-log((x-1)^2)=-2log(x-1)$

Answer 2 · 2017-04-09T17:37:57

Given: $ln(x/(x-1))+ ln((x+1)/x)-ln(x^2-1)$

Addition of logarithms is the same as multiplication of their arguments:

$ln(x/(x-1)(x+1)/x)-ln(x^2-1)$

Simplify a bit:

$ln((x+1)/(x-1))-ln(x^2-1)$

Subtractions of logarithms is the same as division of their arguments:

$ln((x+1)/((x-1)(x^2-1)))$

Factor the quadratic:

$ln((x+1)/((x-1)(x-1)(x+1)))$

Cancel the common factor $x+1$ :

$ln(1/((x-1)(x-1)))$

Write the denominator as a square:

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

Because $1 = 1^2$ , we can write it this way:

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

Use the property $ln(a^c) = (c)ln(a)$ to bring the -2 outside:

$-2ln(x-1)$

Simplify ln(x/(x-1))+ ln((x+1)/x)-ln(x^2-1)ln(x/(x-1))+ ln((x+1)/x)-ln(x^2-1) ?