Estimation of the number of prime numbers in a $b^x$ to $b^{x + 1}$ interval This is a question I have put to myself a long time ago, although only now am I posting it. The thing is, though there is an infinity of prime numbers, they become more and more scarce the further you go.
So back then, I decided to make an inefficient program (the inefficiency was not intended, I just wanted to do it quickly, I took more than 10 minutes to get the numbers below, and got them now from a sheet of paper, not the program) to count primes between bases of different numbers.
These are the numbers I got $($below the first numbers are the exponents of $x$, I used a logarithmic scale, in the first line $\forall \ (n$-- $n+1)$ means $ [b^n; b^n+1[$ $)$
---------0-1--2---3----4----5--6---7---8----9----10--11--12--13-----------------
base  2: |0| 2|  2|   2|   5| 7| 13| 23|  43|  75|137|255|464|
base  3: |1| 3| 13|  13|  31|76|198|520|1380|3741|
base 10: |4|21|143|1061|8363|

I made three histograms from this data (one for each base, with the respective logarithmic scales both on the $x$ and $y$ axes) and drew a line over them, that seemed like a linear function (you can try it yourselves, or if you prefer, insert these into some program like Excel, Geogebra, etc.).
My question is: are these lines really tending (as the base and/or as x grows) to linear or even any kind of function describable by a closed form expression?
 A: The prime number theorem is what you need. A rough statement is that if $\pi(x)$ is the number of primes $p \leq x$, then
$$
\pi(x) \sim \frac{x}{\ln(x)}
$$
Here "$\sim$" denotes "is asymptotically equal to".
A corollary of the prime number theorem is that, for $1\ll y\ll x$, then $\pi(x)-\pi(x-y) \sim y/\ln(x)$. So yes, the number of primes start to thin out for larger $x$; in fact, their density drops logarithmically.
To address your specific question, the PNT implies:
\begin{align}
\pi(b^{x+1}) - \pi(b^x) &\sim \frac{b^{x+1}}{\ln( b^{x+1} )} - \frac{b^x}{\ln( b^x )},
\end{align}
where
\begin{align}
\frac{b^{x+1}}{\ln( b^{x+1} )} - \frac{b^x}{\ln( b^x )} &= \frac{b^{x+1}}{x+1\ln(b)} - \frac{b^x}{x\ln(b)}\\
&= \frac{b^{x+1}}{(x+1)\ln(b)} - \frac{b^x}{x\ln(b)}\\
&=\frac{b^x}{\ln(b)}\left( \frac{b}{x+1} - \frac{1}{x} \right)\\
&=\frac{b^x}{\ln(b)}\left( \frac{ bx-(x+1) }{x(x+1)} \right)\\
&=\frac{b^x}{\ln(b)}\left( \frac{ x(b-1)-1) }{x(x+1)} \right)\\
\end{align}
For $x\gg 1$, we can neglect '$-1$' next to $x(b-1)$ in the numerator and $1$ next to $x$ in the denominator, so that:
\begin{align}
\frac{b^{x+1}}{\ln( b^{x+1} )} - \frac{b^x}{\ln( b^x )} &= \frac{b^x}{\ln(b)}\left( \frac{ x(b-1) }{x^2} \right)\\
&=\frac{b^x(b-1)}{x\ln(b)},
\end{align}
so that
$$
\pi(b^{x+1}) - \pi(b^x) \sim \frac{b^x(b-1)}{x\ln(b)}.
$$
As I'm writing this, I see that this is exactly the same as the answer that @Charles gave.
A: Assuming $b\ge1$ there are
$$
\sim\frac{b-1}{\log b}\cdot\frac{b^x}{x}
$$
primes from $b^x$ to $b^{x+1}$ by the Prime Number Theorem.
