= Solution
If $V_{0x}\ne0$, the first formula in (a) gives $e^{-\gamma t}=1-\gamma x/V_{0x}$ and
$$
t=-\frac1\gamma\ln\left(1-\frac{\gamma x}{V_{0x}}\right).
$$
Substitute this and $1-e^{-\gamma t}=\gamma x/V_{0x}$ into $z(t)$:
$$
\boxed{z(x)=\left(\frac{V_{0z}}{V_{0x}}+\frac{g}{\gamma V_{0x}}\right)x
+\frac{g}{\gamma^2}\ln\left(1-\frac{\gamma x}{V_{0x}}\right)}.
$$
For finite physical times $t\geq0$, the domain is $0\leq\gamma x/V_{0x}<1$, valid for either sign of $V_{0x}$. If $V_{0x}=0$, the trajectory is vertical, $x(t)=0$, and in general no single-valued graph $z(x)$ describes it. As a consistency check, the logarithmic expansion at fixed $x$ cancels the apparent $1/\gamma$ terms and gives $z\to(V_{0z}/V_{0x})x-gx^2/(2V_{0x}^2)$ as $\gamma\to0$, the drag-free <parabola>.
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