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)#