Question #d4b0f

2 Answers
Aritra G.
Sep 8, 2017

We have #y = 2log (x + y) + 2log (x - y) - log (x^2 + y^2)#

#implies 2log [(x+y)(x-y)] - log (x^2 + y^2)#

#implies 2log (x^2 - y^2) - log (x^2 + y^2)#

#implies log [(x^2 - y^2)^2] - log (x^2 + y^2)#

#implies log [(x^2 - y^2)^2/(x^2 + y^2)]#

Taha Z.
Sep 8, 2017

#log(((x+y)^2*(x-y)^2)/(x^2+y^2))#

Explanation:

Before starting the answer, the three basic rules for log combinations are as followed:

  1. #logm +logn = logmn#
  2. #logm - logn = log(m/n)#
  3. #nlogm=log m^n#

Now,
#2log(x+y)+2log(x-y)-log(x^2+y^2)#

First we apply the #3^(rd)# rule to the first and second term:

#log(x+y)^2+log(x-y)^2-log(x^2+y^2)#

Then, we apply the first rule to the first two terms

#log((x+y)^2*(x-y)^2)-log(x^2+y^2)#

Finally the third rule is applied

#log(((x+y)^2*(x-y)^2)/(x^2+y^2))#

I hope that this helps. :)

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