Simplify $ln (\frac{x}{x - 1}) + ln (\frac{x + 1}{x}) - ln (x^{2} - 1)$ $ln(x/(x-1))+ ln((x+1)/x)-ln(x^2-1)$ ?

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Simplify $ln (\frac{x}{x - 1}) + ln (\frac{x + 1}{x}) - ln (x^{2} - 1)$ $ln(x/(x-1))+ ln((x+1)/x)-ln(x^2-1)$ ?

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Answer 1 · 2017-04-09T17:30:43

$- 2 log (x - 1)$ $-2log(x-1)$

Explanation:

Using the logarithmic properties

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

we have

$log (\frac{x}{x - 1}) + log (\frac{x + 1}{x}) - log (x^{2} - 1) = log ⎛ ⎜ ⎝ \frac{(\frac{x}{x - 1}) (\frac{x + 1}{x})}{x^{2} - 1} ⎞ ⎟ ⎠$ $log(x/(x-1))+log((x+1)/x)-log(x^2-1)=log(((x/(x-1))((x+1)/x))/(x^2-1))$

$= log (\frac{\frac{x + 1}{x - 1}}{(x + 1) (x - 1)}) = log (\frac{1}{{(x - 1)}^{2}}) = - log ({(x - 1)}^{2}) = - 2 log (x - 1)$ $=log(((x+1)/(x-1))/((x+1)(x-1)))=log(1/(x-1)^2)=-log((x-1)^2)=-2log(x-1)$

Answer 2 · 2017-04-09T17:37:57

Given: $ln (\frac{x}{x - 1}) + ln (\frac{x + 1}{x}) - ln (x^{2} - 1)$ $ln(x/(x-1))+ ln((x+1)/x)-ln(x^2-1)$

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

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

Simplify a bit:

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

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

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

Factor the quadratic:

$ln (\frac{x + 1}{(x - 1) (x - 1) (x + 1)})$ $ln((x+1)/((x-1)(x-1)(x+1)))$

Cancel the common factor $x + 1$ $x+1$ :

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

Write the denominator as a square:

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

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

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

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

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

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Simplify $ln (\frac{x}{x - 1}) + ln (\frac{x + 1}{x}) - ln (x^{2} - 1)$ $ln(x/(x-1))+ ln((x+1)/x)-ln(x^2-1)$ ?

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Explanation:

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