What is ln(x/(x-1))+ln(((x+1))/x)-ln(x^2-1) ?

The solution to ln(x/(x-1))+ln(((x+1))/x)-ln(x^2-1) is -ln((x-1)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

simplify

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

- ln ((x - 1)^{2})

Solution steps

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

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

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

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

Simplify

\frac{x}{x - 1} \cdot \frac{( x + 1 )}{x} : \frac{x + 1}{x - 1}

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

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

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

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

Simplify

\frac{\frac{x + 1}{x - 1}}{x ^{2} - 1} : \frac{1}{( x - 1 ) ^{2}}

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

Apply log rule:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - ln ((x - 1)^{2})

lo g_{2} (x) + lo g_{2} (x - 2)

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

ln (x) - 2 ln (x^{2} + 1) + \frac{1}{2} ln (3 - x^{4})

lo g_{b} (\frac{p ^{4} q ^{7}}{m ^{5} b ^{7}})

lo g_{10} (y z^{6})

What is ln(x/(x-1))+ln(((x+1))/x)-ln(x^2-1) ?
The solution to ln(x/(x-1))+ln(((x+1))/x)-ln(x^2-1) is -ln((x-1)^2)