Exercise 4.5.13* (There is exactly one power of $2$ having $k$ digits with leading digit $1$)

Question

For any integer $k>1$, prove that there is exactly one power of $2$ having exactly $k$ digits with leading digit $1$, when written in standard fashion to base $10$. For example $2^4 = 16,\ 2^7 = 128$. Prove also that there is exactly one power of $\, 5$ having exactly $k$ digits with leading digit {\bf not} equal to $1$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
\begin{enumerate}
\item[(a)]
If $j$ is a positive integer, $2^j$ has exactly $k$ digits with leading digit $1$ if and only if $10^{k-1} \leq 2^j < 2 \cdot 10^{k-1}.$ Since $2^j$ is never a power of $10$, this is equivalent to
$$10^{k-1} < 2^j < 2 \cdot 10^{k-1}.$$
Moreover,
\begin{align*}
10^{k-1} \l< 2^j < 2 \cdot 10^{k-1} &\iff (k -1)  \ln 10  \leq j \ln 2 < (k-1) \ln 10 + \ln 2\
&\iff (k -1)  \frac{\ln 10}{\ln 2}  < j  < (k-1)  \frac{\ln 10}{\ln 2}  + 1\
&\iff j < (k-1)  \frac{\ln 10}{\ln 2}  + 1 <  j + 1\
&\iff j = 1 + \left \lfloor   (k-1) \frac{\ln 10}{\ln 2} ight floor.
\end{align*}

There is exactly one power of $2$ having exactly $k$ digits with leading digit $1$, given by $2^j$, where
 $$j =  1 +  \left \lfloor   (k-1) \frac{\ln 10}{\ln 2} ight floor.$$
 Example:
 \begin{verbatim}
sage: k = 17
sage: j = 1 + floor((k - 1) * log(10,2)); j
54
sage: 2^j
18014398509481984
 \end{verbatim}
 \item[(b)]
 Similarly, $5^j$ has exactly $k$ digits with leading digit not equal to $1$ if and only if $2\cdot 10^{k-1} < 5^j < 10^k$ (since $2\cdot 10^{k-1}$ is never a power of $5$), and
 \begin{align*}
 2\cdot 10^{k-1} < 5^j < 10^k & \iff (k-1) \ln 10 + \ln 2 < j \ln 5 < k \ln 10\
 &\iff k \ln 10 - \ln 5 < j \ln 5 < k \ln 10\
 &\iff j \ln 5 < k \ln 10 < (j+1) \ln 5\
 &\iff  j  < k\, \frac{ \ln 10}{\ln 5} < j + 1\
 &\iff j = \left \lfloor   k \, \frac{\ln 10}{\ln 5} ight floor
 \end{align*}
 There is exactly one power of $5$ having exactly $k$ digits with leading digit not equal to $1$, given by $5^j$, where
 $$j = \left \lfloor   k \, \frac{\ln 10}{\ln 5} ight floor.$$
Example:
\begin{verbatim}
sage: k = 5
sage:  j = floor( k * log(10, 5))
sage: 5^j
78125
\end{verbatim}
 \end{enumerate}

\end{proof}

Exercise 4.5.13* (There is exactly one power of $2$ having $k$ digits with leading digit $1$)

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Exercise 4.5.13* (There is exactly one power of $2$ having $k$ digits with leading digit $1$)

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