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

Question

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

Answer 1

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: $$n=50,x=1$$ $$n=50,x=1.096...$$ $$n=50,x=1.19347...$$ $$n=50,x=1.29232...$$ $$n=50,x=1.39248...$$

Answer 2

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.