How to solve #log(x-1)+log(x+1)=2log(x+2)#?

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Answer 1 · 2018-03-01T14:09:29

For #x>1# we have that

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

#Log(x^2-1)=log(x+2)^2#

#x^2-1=(x+2)^2#

#x^2-1=x^2+4x+4#

#x=-5/4#

However this solution is not acceptable because we are looking solutions for #x>1#.That is because #loga# is defined for #a>0#

Hence there are NO solutions for this equation.

The following logarithm rules were used

#loga+logb=log(ab)#

#bloga=log(a^b)#

If #loga=logb#, then #a=b#