The prime number theorem gives an increasingly (proportionally) accurate approximation for the number of primes less than a given integer $n$.
Can we use this to find an equally accurate approximation which maps the $k$th prime number to its value $p$?
On the x axis should be the number of primes below $n$, $\frac{n}{log(n)}$ and the y axis the value of the number, $p$. So this is the inverse function for $\frac{n}{log(n)}$.
So would the value of the $k$th prime number be approximately $p \approx \frac{k}{\log k}$?
Formula's using any principle other than the PNT are also welcome.
Well, the Prime Number Theorem says that, where $g(x)=\frac{x}{\ln x}$, the $g(N)$th prime is approximately $N$. So the function you are looking for is $f(x)=g^{-1}(x)$. I don't know of any easy way to find the inverse of $g(x)$.
Asymptotically, however, we have $f(x)=O(x \ln x)$. This can be seen by letting $f(x)=x (\ln x + h(x))$, for some function $h$. We know that $x=f(\frac{x}{\ln x})$; substituting in the expression for $f$ yields $x=\frac{x}{\ln x}\left(\ln\frac{x}{\ln x} + h\left(\frac{x}{\ln x}\right)\right) = \frac{x}{\ln x}\left(\ln{x} - \ln \ln x + h\left(\frac{x}{\ln x}\right)\right)$. When this is simplified, we get that $h\left(\frac{x}{\ln x}\right) = \ln \ln x$. It should be clear now that the growth of $h(x)$ is slower than that of $\ln x$ (you can see this by plugging in $h(x)=c\ln x$ and getting that $h$ grows too fast to satisfy the equality). Thus, $f(x)=x (\ln x + h(x))$ is $O(x \ln x)$.