Exponential: proof/explanation for Int[dx/x]=ln x
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From: Stefan.Duke on 19 Jul 2010 10:41 Hi, there is one thing that puzzles me in the derivation of exponential growth. Let's start with dx/dt = kx where t is time and the rate. Then dx/x= k dt and then Int[dx/x]=Int[k dt] ln x = kt + constant Can somebody explain or direct me to a source explain why Int[dx/x]=ln x ? Thanks and best, stefan
From: Greg Neill on 19 Jul 2010 10:59 Stefan.Duke(a)gmail.com wrote: > Hi, > there is one thing that puzzles me in the derivation of exponential > growth. > Let's start with > dx/dt = kx > > where t is time and the rate. > Then > dx/x= k dt > and then > Int[dx/x]=Int[k dt] > ln x = kt + constant > > Can somebody explain or direct me to a source explain why Int[dx/x]=ln > x ? > Thanks and best, > stefan Are you aware that d(e^x)/dx = e^x ? Consider the relationship y = e^x. If we differentiate both sides: dy/dx = e^x = y so we can write: dy/y = dx Integrating both sides: INT(dy/y) = x + c but from above, y = e^x, so taking the natural log of each side, ln(y) = x. Thus INT(dy/y) = ln(y) + c
From: cbb on 19 Jul 2010 11:40 On Jul 19, 4:59 pm, "Greg Neill" <gneil...(a)MOVEsympatico.ca> wrote: > Stefan.D...(a)gmail.com wrote: > > Hi, > > there is one thing that puzzles me in the derivation of exponential > > growth. > > Let's start with > > dx/dt = kx > > > where t is time and the rate. > > Then > > dx/x= k dt > > and then > > Int[dx/x]=Int[k dt] > > ln x = kt + constant > > > Can somebody explain or direct me to a source explain why Int[dx/x]=ln > > x ? > > Thanks and best, > > stefan > > Are you aware that d(e^x)/dx = e^x ? > > Consider the relationship y = e^x. If we differentiate > both sides: > > dy/dx = e^x = y > > so we can write: > > dy/y = dx > > Integrating both sides: > > INT(dy/y) = x + c > > but from above, y = e^x, so taking the natural log of each > side, ln(y) = x. Thus > > INT(dy/y) = ln(y) + c thanks for the quick response. So if I get it right, this only holds when I define Y=exp(x) (given some scale parameter for the growth rate). It wouldn't hold for any other function, say, y=x^2?
From: Alois Steindl on 19 Jul 2010 11:43 cbb <stefan.lhachimi(a)gmail.com> writes: > thanks for the quick response. So if I get it right, this only holds > when I define Y=exp(x) (given some scale parameter for the growth > rate). It wouldn't hold for any other function, say, y=x^2? Can't you just try it out? Alois
From: Stefan.Duke on 19 Jul 2010 15:17
On Jul 19, 5:43 pm, Alois Steindl <Alois.Stei...(a)tuwien.ac.at> wrote: > cbb <stefan.lhach...(a)gmail.com> writes: > > thanks for the quick response. So if I get it right, this only holds > > when I define Y=exp(x) (given some scale parameter for the growth > > rate). It wouldn't hold for any other function, say, y=x^2? > > Can't you just try it out? > > Alois I did and it didn't hold. But that doesn't mean that I am right, I might have just overlooked something ;) |