How do you condense #4log(x)-2log((x^2)+1)+2log(x-1) #?

1 Answer
Apr 8, 2016

Answer:

#log((x^4(x -1)^2)/(x^2+1)^2) = log((x^6 - 2x^5 + x^4)/(x^4 + 2x^2 + 1))#

Explanation:

First, get rid of all the coefficients for the logarithms

#4logx = logx^4#
#-2log(x^2 + 1) = log(x^2 + 1)^(-2)#
#2log(x-1) = log(x-1)^2#

Now you can rewrite the equation above as

#4logx - 2log(x^2 + 1) + 2log(x-1)#
# = logx^4 + log(x^2 + 1)^(-2) + log(x-1)^2#

Finally, knowing that adding together #log#s is the same as having one #log# with all of the #x# bits multiplied together inside,

#logx^4 + log(x^2 + 1)^(-2) + log(x-1)^2 #
# = log(x^4(x^2+1)^-2(x-1)^2)#
# = log((x^4(x-1)^2)/(x^2+1)^2)#

You can expand this out if necessary, to

#log((x^6 - 2x^5 + x^4)/(x^4 + 2x^2 + 1))#