Solve $\pi(xn)=\frac{xn}{\frac{n}{\pi(n)}+\ln(x)}$ 
Solve $$\pi(xn)=\frac{xn}{\frac{n}{\pi(n)}+\ln(x)}.$$

By the prime number theorem, we have:
$$\pi(n)\sim\frac{n}{\ln(n)}$$
Therefore: 
$$\pi(xy)\sim\frac{xy}{\ln(xy)}$$
$$\pi(xy)\sim\frac{xy}{\ln(x)+\ln(y)}$$
$$\frac{xy}{\pi(xy)}\sim \ln(x)+\ln(y)$$
$$\frac{xy}{\pi(xy)}-\ln(x)\sim \ln(y)$$
$$\frac{y}{\frac{xy}{\pi(xy)}-\ln(x)}\sim\frac{y}{\ln(y)}$$
$$\frac{y}{\frac{xy}{\pi(xy)}-\ln(x)}\sim\pi(y)$$
$$\pi(xy)\sim\frac{xy}{\frac{y}{\pi(y)}+\ln(x)}$$
We get an asymptotic relation between $\pi(xy)$ and $\pi(y)$. Now since the prime counting function is defined on $\mathbb {R}$, how can I find, for any integer $n$, the value $x$ such that:
$$\pi(xn)=\frac{xn}{\frac{n}{\pi(n)}+\ln(x)}$$
 A: This is too long for a comment.
Having problems with $\pi(xn)$ when $x$ is a real, nevertheless, I considered looking to the zero's of  function
$$f(x)=\pi ( n x)-\frac{n x}{\frac{n}{\pi (n)}+\log (x)}$$ which was accepted by a CAS.
For $n=23$, the first roots of $f(x)=0$ with $x>1$ are $1.343$, $1.378$, $1.865$, $1.966$, $2.043$, $2.167$, $2.380$, $2.565$, $2.593$, $2.652$ and so on; the last I found being $4.390$. 
This makes a bunch of solutions.
A: Ok so the question did not mention it, but there is the solution x=1, for any n. Now let's take n=50. The first solution is $n=50$, $x=1$:
$$\pi(50)=\frac{50}{\frac{50}{\pi(50)}+\ln(1)}=15$$
Now as x grows, we will also get:
$$\pi(50*\frac{53}{50})=\frac{50*\frac{53}{50}}{\frac{50}{\pi(50)}+\ln(\frac{53}{50})}\approx16$$
$$\pi(50*\frac{59}{50})=\frac{50*\frac{59}{50}}{\frac{50}{\pi(50)}+\ln(\frac{59}{50})}\approx17$$
$$\pi(50*\frac{61}{50})=\frac{50*\frac{61}{50}}{\frac{50}{\pi(50)}+\ln(\frac{61}{50})}\approx18$$
$$\pi(50*\frac{67}{50})=\frac{50*\frac{67}{50}}{\frac{50}{\pi(50)}+\ln(\frac{67}{50})}\approx19$$
After that, it will not be possible because: 
$$\pi(xn)-\frac{xn}{\frac{n}{\pi(n)}+\ln(x)}>1$$
And this difference will grow as x grows.
So now we can find x for $\pi(xn)=16,17,18,19$ using the product log function:
$$ x=\frac{-16W\left(-\frac{ne^{-\frac{n}{\pi(n)}}}{16}\right)}{n} $$
$$ x=\frac{-17W\left(-\frac{ne^{-\frac{n}{\pi(n)}}}{17}\right)}{n} $$
$$ x=\frac{-18W\left(-\frac{ne^{-\frac{n}{\pi(n)}}}{18}\right)}{n} $$
$$ x=\frac{-19W\left(-\frac{ne^{-\frac{n}{\pi(n)}}}{19}\right)}{n} $$
(from Wolfram Alpha)

Finally, we get the following solutions:
$$n=50,x=1$$
$$n=50,x=1.096...$$
$$n=50,x=1.19347...$$
$$n=50,x=1.29232...$$
$$n=50,x=1.39248...$$
