# Simplifying log eqns

1. Jan 26, 2007

### Bazman

Hi,

To solve the following d.e.:

xdy-ydx = 0 1

you get

ln(y)-ln(x)=C 2

no clearly you can simplify to:

ln(y/x)=c 3

which after taking exponentials gives:

y=Ax where A=e^c 4

however what interests me is if you do not simplify to

ln(y/c) in line 3 but simply take the exponential of line 2

then you get

y-x=A

obviously this is incorrect but I would like to know where I have gone wrong in the step above and also if it possible to get to the correct answer without using the simplification in step 3 above

2. Jan 26, 2007

### cristo

Staff Emeritus
Equation (2) reads lny-lnx=C. Taking exponential of both sides yields e(lny-lnx)=A. Simplifying this gives elnye-lnx=elnyelnx-1=yx-1.

Your mistake was saying that e(lny-lnx)=y-x, which is not true.

3. Jan 26, 2007

### HallsofIvy

$e^{a- b}$ is $e^a/e^b$, not a- b so $e^{lna- ln b}$ is NOT a- b
$$e^{ln a- ln b}= e^{ln \frac{a}{b}}= \frac{a}{b}$$