I have to generate a graph, with $y = e^{a\cdot t} - 1$ and $x = e^{b\cdot t} - 1 $ for all possible values of t. Constants a & b are given. In other words, the curve generated will be the locus of all points having all possible values of t. Thanks for your help!
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Would this include strange values of t like $i$? There is the question of what is the domain of t as you don't state whether t could be complex or not. – JB King Mar 20 '13 at 18:28
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I will assume that $a,b$ are real constants and $t$ is a real variable.
You could invert the formula for $x$, to get $$t = \frac {\log (x+1)} b$$ and then plug it in the formula for $y$, to get $$y = \exp\left({\frac a b \log (x+1)}\right) - 1 = (x+1)^{a/b} - 1$$
The domain for this formula is $x > -1$.
Yoni Rozenshein
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