How do you solve #ln (x + 2) + ln (x - 2) = 0#?

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Answer 1 · 2016-06-23T09:32:43

#x=sqrt5#

#ln(x+2)+ln(x-2)=0# is equivalent to

#ln[(x+2)(x-2)]=ln1#

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

or #x^2-4=1#

or #x^2-5=0#

or #(x+sqrt5)(x-sqrt5)=0#

or #x=-sqrt5# or #x=sqrt5#

But as #x=-sqrt5# is not in domain (as log of negative number is not possible) hence #x=sqrt5#