I want to estimate the amount of prime numbers up to $\sqrt{6657\cdot 2^{10000}+1}$ by manual calculation.
My attempt is as follows: Due to the prime number theorem we know that $\frac{x}{\log(x)}$ is a good enough estimation in this range. Wolframalpha told me, that $\sqrt{6657\cdot 2^{10000}+1} \approx 1.153\cdot 10^{1507}$. So I tried to estimate $ \pi(1.1\cdot 10^{1507}) $ as follows:
$ \pi(1.1\cdot 10^{1507}) \approx \frac{1.1\cdot 10^{1507}}{\log(1.1\cdot 10^{1507})} = \frac{11}{10}\cdot 10^{1507} \cdot \frac{1}{\log(\frac{11}{10}\cdot 10^{1507})} = \frac{11}{10}\cdot 10^{1507} \cdot \frac{1}{\log(11)-\log(10) + 1507\cdot\log(10)}. $
Now let's look at the fraction: $$ \frac{1}{\log(11)-\log(10) + 1507\cdot\log(10)} \geq \frac{1}{\log(11) + 1507\cdot\log(10)} \geq \frac{1}{3 + 1507\cdot 3} = \frac{1}{9045} > \frac{1}{10000}. $$
So I get: $$ \frac{11}{10}\cdot 10^{1507} \cdot \frac{1}{\log(11)-\log(10) + 1507\cdot\log(10)} > \frac{11}{10}\cdot 10^{1507}\cdot \frac{1}{10000} > 10\cdot 10^{1507}\cdot 10^{-3} = 10^{1505}. $$
Are my estimations correct?
I found my mistake thanks to @Aphelli:
It is $$ \begin{align} &\frac{11}{10}\cdot 10^{1507}\cdot\frac{1}{\log(11) - \log(10)+1507\log(10)} \\ > & 1\cdot 10^{1507}\frac{1}{\log(11)+1507\log(10)} \\ > & 10^{1507} \cdot \frac{1}{3+1507\cdot 3} \\ = & 10^{1507} \cdot \frac{1}{4524} \\ > & 10^{1507} \cdot \frac{1}{5000} \\ = & 10^{1507} \cdot \frac{1}{5}\cdot 10^{-3} \\ = & \frac{1}{5}\cdot 10^{1504} \\ = & 2\cdot 10^{1503}, \end{align} $$ what is a very good estimation.