Solving Log Equations: No Simplification Needed

[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

 

[cristo]

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

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

 

[HallsofIvy]

[itex]e^{a- b}[/itex] is [itex]e^a/e^b[/itex], not a- b so [itex]e^{lna- ln b}[/itex] is NOT a- b
[tex]e^{ln a- ln b}= e^{ln \frac{a}{b}}= \frac{a}{b}[/tex]