I will answer your question "...so that I can make $x$ the subject?", but first, you have an error.
In moving from line 2 to line 3, you apply the natural log to both sides. However, you cannot use $\ln$ the way you do.
I will change your equation to make the example easier to follow.
Line 2 is something like $\frac{y}{14}=1-x^{19}$.
When applying $\ln$ to both sides, we must apply it to both sides.
Line 3 should be like $\ln\left(\frac{y}{14}\right)=\ln\left(1-x^{19}\right)$. In your work, you didn't have the $1$ in the argument of $\ln$.
So, what to do? We need to isolate the $x$-term before using $\ln$.
Working from Line 2...
\begin{align}
\frac{y}{145366.25}&=1-\left(\frac{x}{1013.25}\right)^{0.190284}\\
\frac{y}{145366.25}-1&=-\left(\frac{x}{1013.25}\right)^{0.190284}&\text{Isolate $x$-term}\\
1-\frac{y}{145366.25}&=\left(\frac{x}{1013.25}\right)^{0.190284}&\text{Divide by $-1$}\\
\ln\left(1-\frac{y}{145366.25}\right)&=\ln\left[\left(\frac{x}{1013.25}\right)^{0.190284}\right]&\text{Apply $\ln$ to both sides}\\
\ln\left(1-\frac{y}{145366.25}\right)&=0.190284\ln\left(\frac{x}{1013.25}\right)&\text{Power Rule for $\ln$}\\
\frac{\ln\left(1-\frac{y}{145366.25}\right)}{0.190284}&=\ln\left(\frac{x}{1013.25}\right)
\end{align}
Now, what to do with $\ln\left(\frac{x}{1013.25}\right)$? We use the Quotient Rule for Logarithms.
$$\ln\left(\frac{a}{b}\right)=\ln a-\ln b$$
Picking up where we stopped...
\begin{align}
\frac{\ln\left(1-\frac{y}{145366.25}\right)}{0.190284}&=\ln\left(\frac{x}{1013.25}\right)\\
\frac{\ln\left(1-\frac{y}{145366.25}\right)}{0.190284}&=\ln x -\ln 1013.25&\text{Quotient Rule for Logs}\\
\frac{\ln\left(1-\frac{y}{145366.25}\right)}{0.190284}+\ln 1013.25&=\ln x
\end{align}
Raise both sides to base $e$, or use the inverse property of the natural log, and you have $x$ isolated.
