#log(x^2+4)-log(x+2) = 2+log(x-2)#
Let us try to get this as #log(A) = log(B)# so that we can remove the logs and equate #A=B#
#2 = 2log(10) = log(10^2) = log(100)#
Our equation would become
#log(x^2+4)-log(x+2) = log(100)+log(x-2)#
#log((x^2+4)/(x+2))=log(100(x-2))#
Rule: #log(A)-log(B) = log(A/B)# and #log (A) + log(B) = log(AB)#
#(x^2+4)/(x+2) = 100(x-2) #
#(x^2+4) = 100(x-2)(x+2)# this is got by multiplying both sides by (x+2)
#x^2+4 = 100(x^2-4)#
#x^2+4 = 100x^2 - 400#
Add #400# to both the sides.
#x^2+404 = 100x^2#
Subtract x^2 from both the sides.
#404 = 99x^2#
Divide by 99 on both the sides to isolate x^2
#x^2 = 404/99#
#x=+-sqrt(404/99)#
#x=sqrt(404/99)# is the solution
#x=-sqrt(404/99)# does not satisfy the original equation and is discarded.
#x=sqrt(404/99)# Answer
Note: if you want an approximate value, use calculator and round off to the required decimal places.
