How to Get Rid of Natural Logarithms in Separable Differential Equations

[Aerosion]

Homework Statement

so here's my equation:

dy/dx=(xy+3x-y-3)/(xy-2x+4y-8)

so what i did first was factor out the right side

=(x+1)(y-3)/(x+4)(y-2)

then i did a bunch of manipulation to get the ys on one side and the xs on another (i won't write this out right now but if anyone wants me to i can)

and got

(y-2)/(y+3) dy = (x-1)/(x+4) dx

of course i integrate, i get

y-5*ln(y+3) = x-5ln(x+4)

i want to get rid of the lns, right? so i multiplied them by e (like e^(ln(y+3)

then i got -4y-15=-4x-20

i'm not sure what to do after this because i looked at the answer at the back of my book and it said something completely different to what I've done so far, any help?

Homework Equations

The Attempt at a Solution

 

[gabbagabbahey]

Shouldn't your exponent for $e$ be the entire expression on each side of the equation?

i.e.

$$e^{y-5ln(y+3)}=e^{x-5ln(x+4)}$$