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Calculus1
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ln(y) = -2x + c Just did separation of variables. Solving for y (general solution). I cannot for the life of me figure out how to undo a natural log.
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e^(ln(y))=y for y>0
\[y=e^{-2x+c}\] you can make answer look prettier
Woah. e^lny = y? Man, I need to relearn natural log stuff. That's not connecting for me at all
yes f(x)=ln(x) has inverse function g(x)=e^x ans so f((g(x))=x and g(f((x))=x on their domains
\[y=ke^{-2x} \text{ since } e^c \text{ is a constant }\]
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is what I meant about making the answer look prettier
*Whistles* Thanks
lol you whistled at a problem
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