Let \(d(n)\) be the number of digits of the decimal expansion of \(2^n\text{.}\) From our discussion above, we know that \(d(n)\) is one more than the integer part of \(\log 2^n\text{.}\) Using the change of base formula between from base-\(10\) to base-\(2\text{,}\) we have
\begin{align*}
\log 2^n \amp = \log_{10}2^n \\
\amp = \frac{\log_2 2^n}{\log_2 10}\\
\amp = \frac{n}{\log_2 10}\text{.}
\end{align*}
Since \(3\leq \log_2 10\leq 4\text{,}\) we have
\begin{align*}
\frac{n}{4} \amp \leq \log 2^n=\frac{n}{\log_2 10}\leq \frac{n}{3} \text{.}
\end{align*}
Putting it all together, we see that
\begin{align*}
\frac{n}{4}\amp \leq d(n) \leq \frac{n}{3}+1
\end{align*}
In particular, we estimate that \(2^{10}\) has between \(10/4=2.5\) and \(10/3+1\approx 4.3\) digits. More precisely, since the number of digits is an integer, \(2^{10}\) should have between \(3\) and \(4\) digits. In fact, we have
\begin{align*}
2^{10} \amp =(2^{5})^2\\
\amp =(32)^2\\
\amp = 1024\text{.}
\end{align*}
Our estimate is correct!
Similarly, since \(50/4=12.5\) and \(50/3\approx 16.7\text{,}\) \(2^{50}\) should have between \(13\) and \(16\) digits. In fact, using technology, we find that
\begin{align*}
2^{50} \amp = 1,125,899,906,842,624 \text{,}
\end{align*}
which has \(16\) digits. Again, our estimate was correct!
You may have noticed that the upper bound of estimates yielded for
\(d(n)\) happened to be equal to the actual answer. This is just a happy accident for the values of
\(n\) we chose. For example,
\(2^{12}=4096\) has
\(4\) digits, whereas our estimation method predicts between
\(3\) and
\(5\) digits.
