Q:

Consider the following: $x = e^{4t}$, $ y = t + 3$ Eliminate the parameter to find a Cartesian equation of the curve.

Consider: $x = e^{4t}$,   $ y = t + 3$
(a) Eliminate the parameter to find a Cartesian equation of the curve.
My answer: 
Disregard the one and replace it with three $y=3+\frac{ln(x)}{4}$
(b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. 
I thought ln(x) was not defined for negative values? 

A:

You've eliminated the bottom two graphs because $\ln x$ is not defined for $x<0$. Now all you have to do is identify the difference between the top two.
The arrow denotes how $x$ and $y$ change with $t$. In the first graph, the arrow tells you that, as $t$ increases,

$x$ increases $\left(\frac{\mathrm{d}x}{\mathrm{d}t}>0\right)$.
$y$ increases $\left(\frac{\mathrm{d}y}{\mathrm{d}t}>0\right)$.

In the second graph, the arrow tells you that, as $t$ increases,

$x$ decreases.
$y$ decreases.

Only one of these combinations is true. Now look at the derivatives of your original two parametric equations:
$$\frac{\mathrm{d}x}{\mathrm{d}t}=4e^{4t}>0,\quad\frac{\mathrm{d}y}{\mathrm{d}t}=1>0$$
This should tell you which of the two graphs is correct. Actually, it differentiates (pun intended) all four graphs from one another.
I suspect, however, that there has been an error in drawing the graphs, because you clearly shouldn't get some finite value for $\ln 0$ - the result of
$$\lim_{x\to0}\ln x$$
is clearly $-\infty$! It may be that the screen resolution isn't good enough, given that the graphs are zoomed out quite a lot.

