#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.