exponential functions

katyt

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Sep 7, 2005
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this might seem like a stupid question, but I cannot remember for the life of me the rules for this. ok so i'm solving a differential equation and i end up with:
s=e^(.5ln(2t)+t^2+C)
I'm having trouble remembering the rules for exponential functions. I can't figure out how to rewrite/simplify this. Any help?
Thanks! Katy
 
Hello, Katy!

That was not a stupid question . . . This is a tricky one.

i'm solving a differential equation and i end up with: .s=e[0.5ln(2t)+t2+c]\displaystyle s\:=\:e^{[0.5\ln(2t)\,+\,t^2\,+\,c]}

I can't figure out how to rewrite/simplify this.
Recall that: .ea+b=eaeb\displaystyle e^{a+b}\:=\:e^a\,\cdot\,e^b

So: .e[0.5ln(2t)+t2+c]  =  e0.5ln(2t)et2ec\displaystyle e^{[0.5\ln(2t)\,+\,t^2\,+\,c]}\;=\;e^{0.5\ln(2t)}\,\cdot\,e^{t^2}\,\cdot\,e^c

Since ec\displaystyle e^c is just another constant C\displaystyle C, we have: .Cet2e0.5ln(2t)\displaystyle C\cdot e^{t^2}\cdot e^{0.5\ln(2t)}

Then: .Cet2eln(2t)12=Cet2(2t)12=Cet22t\displaystyle C\cdot e^{t^2}\cdot e^{\ln(2t)^{\frac{1}{2}}}\:=\:C\cdot e^{t^2}\cdot(2t)^{\frac{1}{2}}\:=\:C\cdot e^{t^2}\cdot\sqrt{2t}
 
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