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

2 Answers
Apr 24, 2017

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

Explanation:

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)

Apr 24, 2017

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

Explanation:

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)