Problem source

To nine decimal places, $\log_{10}2=0.301029996$ and
 $\log_{10}3=0.477121255$.
\begin{questionparts}
\item  Calculate $\log_{10}5$ and  $\log_{10}6$ to three decimal
places. By taking logs, or otherwise,  show that 
\[
5\times 10^{47} < 3^{100} < 6\times 10^{47}.
\]
Hence write down the first digit of $3^{100}$.
\item  Find the first digit of each of the following numbers:
$2^{1000}$; \ $2^{10\,000}$; \ and 
$2^{100\, 000}$.
\end{questionparts}

Solution source

\begin{questionparts}
\item 
\begin{align*}
\log_{10}5 &= \log_{10} 10 - \log_{10}2 \\
&= 1- \log_{10} 2 \\
&= 0.699\\
\\
\log_{10} 6 &= \log_{10} 2 + \log_{10} 3 \\
&= 0.301029996+0.477121255 \\
&= 0.778
\end{align*}

\begin{align*}
&& 5 \times 10^{47} < 3^{100} < 6 \times 10^{47} \\
\Leftrightarrow && 47 + \log_{10} 5 < 100 \log_{10} 3 < \log_{10} 6 + 47 \\ 
\Leftrightarrow &&47.699< 47.71 < 47.778 \\ 
\end{align*}
Which is true. Therefore the first digit of $3^{100}$ is 5.
\item $\log_{10} 2^{1000} = 1000 \log_{10} 2 = 301.02\cdots$. Therefore it starts with a $1$.

$\log_{10}2^{10\, 000} = 10\,000 \log_{10} 2 = 3010.2$ therefore this also starts with a $1$.

$\log_{10} 2^{100\, 000} = 100\,000 \log_{10} 2 = 30102.9996$ therefore it starts with a $9$
\end{questionparts}