How do you condense #2(3lnx-ln(x+1)-ln(x-1))#?

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Answer 1 · 2017-04-24T08:30:58

Depends how far you want to take things but as a single logarithm it becomes #ln((x^3(x-1))/(x+1))^2#

Multiples of logarithms become powers:
#2(3ln(x)-ln(x+1)-ln(x-1))#
#2(ln(x^3)-ln(x+1)-ln(x-1))#

Subtracting logarithms is equivalent to dividing their arguments:
#2(ln((x^3)/(x+1))-ln(x-1))#

Now divide again:
#2ln(x^3/((x+1)(x-1)))#

Tidy this up to give:
#2ln((x^3)/(x^2-1))#

You can apply the power law again:
#ln((x^6)/(x^2-1)^2)#

Answer 2 · 2017-04-24T08:48:58

#ln(x^6/(x^2-1)^2)#

Expression #= 2(3ln(x) - ln(x+1) -ln(x-1))#

Here we will use three properties of logarithms:

(i): #alnx = lnx^a#

(ii): #lna + lnb= ln(ab)#

(iii: #lna - lnb -=ln(a/b)#

First let's rewite the Expression as:

= #2(3ln(x) - (ln(x+1) +ln(x-1)))#

Using property (i):

#= 2(lnx^3 - (ln(x+1) +ln(x-1)))#

Using properties (ii) and (iii):

#= 2*lnx^3/(ln((x+1)(x-1))#

#= 2* lnx^3/(ln(x^2-1)#

Using property (i) again:

#= ln(x^3/((x^2-1)))^2#

#= ln(x^6/(x^2-1)^2)#