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.
