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$\begingroup$ $$x=4 \ln(t), \qquad y=2t$$
Write it in Cartesian form, giving $y$ explicitly in terms of $x$.
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$\begingroup$ What you need to do is to eliminate $t$ from both expressions (or equivalently, write $t=f(x)$ or $t=f(y)$ and substitute into the other expression).
In this case, $t=y/2$, so we have $$x=4\ln(y/2),\mbox{ or }y=2\exp((1/4) x),$$ or $t=\exp((1/4) x)$, so $$y=2\exp((1/4) x).$$
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