Number Theory problem

We seek $2^{n} \approx 10^{x}$ but also $2^{n} < 10^{x}$ . The approximation should be very very close.

Using a base 2 logarithm we find, $n < x \cdot \log_{2}10$ . Or $\log_{2}10 > \frac{n}{x}$

We seek a rational approximation for $\log_{2}10$ that is ever so slightly too small.

Fortunately, there is an On-Line Encyclopedia of Integer Sequences with just what we need: Numerators of convergents to log_2(10) Our $n$ is somewhere on this list.

The problem now is which one ? I basically just tried them one, by one in this high precision calculator until I got there.

To 100 decimal places, The input: 10^(frac(1923400330*log(n(2,100),10)))

And the result: 9.999999999721382843735927151859775967059640108039803362830811802931061843671814991066226503

Means $\large n=\boxed{1923400330}$

Which is actually ten 9's in a row, but the next best choice only has eight. But because this is a convergent, there can't be a better choice.

The actual approximation is $2^{1923400330} \approx 9.99999999972138284373592715185977596705964010803980336283081180293106184367181488\times 10^{579001192}$