Simplify ln(x/(x-1))+ ln((x+1)/x)-ln(x^2-1)ln(xx−1)+ln(x+1x)−ln(x2−1) ?

2 Answers
Cesareo R.
Apr 9, 2017

-2log(x-1)−2log(x−1)

Explanation:

Using the logarithmic properties

log(a b)=loga + log blog(ab)=loga+logb
loga^b = b log alogab=bloga

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(xx−1)+log(x+1x)−log(x2−1)=log⎛⎜⎝(xx−1)(x+1x)x2−1⎞⎟⎠

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

Douglas K.
Apr 9, 2017

Given: ln(x/(x-1))+ ln((x+1)/x)-ln(x^2-1)ln(xx−1)+ln(x+1x)−ln(x2−1)

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

ln(x/(x-1)(x+1)/x)-ln(x^2-1)ln(xx−1x+1x)−ln(x2−1)

Simplify a bit:

ln((x+1)/(x-1))-ln(x^2-1)ln(x+1x−1)−ln(x2−1)

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

ln((x+1)/((x-1)(x^2-1)))ln(x+1(x−1)(x2−1))

Factor the quadratic:

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

Cancel the common factor x+1x+1:

ln(1/((x-1)(x-1)))ln(1(x−1)(x−1))

Write the denominator as a square:

ln(1/(x-1)^2)ln(1(x−1)2)

Because 1 = 1^21=12, we can write it this way:

ln((1/(x-1))^2)ln((1x−1)2)

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

-2ln(x-1)−2ln(x−1)

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