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

Question

2463 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-17T11:57:10

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

Condense the last #log# by recognising that a coefficient outside of the log is the same as an exponent inside, or

#3lnx=lnx^3#.

Using the law that #loga+logb=logab#, and #loga-logb=log(a/b)#, then,

#ln(x+1)+ln(x-1)=ln[(x+1)(x-1)]#
#=ln(x^2-1)#

and

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

which you could also condense, using laws of exponents, into

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