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How to graph the parametric equation: $y=e^{-t}$ and $x=e^{t}$

I preferable would like to cancel out the t's on both sides so then I instinctively ln both sides but that gives me ln(x)=t and ln(y)=-t to devolve into me adding the two equations to finally devolve into the following:

x=y which doesn't seem like the real solution/equation to the problem

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    $\begingroup$ Notice that $x y=1$ $\endgroup$ – Claude Leibovici Feb 22 '17 at 7:45
  • $\begingroup$ Adding the equations gives $\ln(x)= -\ln(y), $ not $\ln(x)=\ln(y).$ Then this can be solved to give what Claude says $\endgroup$ – spaceisdarkgreen Feb 22 '17 at 7:55
  • $\begingroup$ Also note that $x$ and $y$ have to be positive. So the correct graph is only the 1st-quadrant part of $y = 1/x.$ $\endgroup$ – B. Goddard Feb 22 '17 at 12:00
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$y = e^{-t} \implies y = \frac{1}{e^t} = \frac{1}{x} $ as $x = e^t$. So the equation is is $y = \frac{1}{x}$.
The way you have solved it (by taking ln) gives $ln(x) +ln(y) =0 \implies ln(xy) = 0 $ or $xy =1$.

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In your substitution that gave you $x=y$ you seem to have made a mistake. Here is a correct substitution, $$x = e^t\ \ \implies\ \ \ln(x) = t$$ $$\therefore\ \ y=e^{-t}=e^{-\ln(x)}=e^{\ln{\frac{1}{x}}}=\frac{1}{x}$$

Or as stated in the comments, $$xy = 1$$ which was kinda obvious if you just multiplied $x$ and $y$ from the start.

Your intuition to use $\ln$ was good, but remember that $$-\ln{z}=\ln{\frac{1}{z}}$$ and more importantly that, $$e^{-\ln(z)} \neq -z$$

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