Approximation to Power of 7 by Power of 10

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Theorem

$7^{510} \approx 1 \cdotp 00000 \, 09377 \, 76536 \ldots \times 10^{431}$

This is the closest known approximation of a power of $7$ by a power of $10$.


Proof

An intuition is given as follows:


Suppose for some $m, n \in \N$:

$7^m = c \cdot 10^n$, where $c$ is very close to $1$.

Taking common logarithm:

$m \log 7 = \log c + n$

Which leads to:

$\log 7 = \dfrac n m + \dfrac {\log c} m$

where $\dfrac {\log c} m$ is very close to $0$.

To make a good approximation is to minimize $\size {\log c}$.


One may calculate the continued fraction of $\log 7$.

We use continued fractions because Convergents are Best Approximations.

One obtains $\sqbrk {0; 1, 5, 2, 5, 6, 1, 4813, 1, 1, 2, 2, \dots}$.

This sequence is A082571 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Let $\dfrac {p_n} {q_n}$ be the convergent $\sqbrk {0; a_1, \dots, a_n}$.

Then:

\(\ds \size {\log c}\) \(=\) \(\ds q_k \size {\log 7 - \frac {p_k} {q_k} }\)
\(\ds \) \(<\) \(\ds q_k \paren {\frac 1 {q_k q_{k + 1} } }\) Accuracy of Convergents of Continued Fraction
\(\ds \) \(=\) \(\ds \frac 1 {q_{k + 1} }\)
\(\ds \) \(=\) \(\ds \frac 1 {a_{k + 1} q_k + q_{k - 1} }\) Recursive Definition of Denominators of Convergents

Thus, at least intuitively, the larger the value of the next $a_i$, the smaller $\size {\log c}$ is.


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In particular: I am not familiar with this, this is just some justification
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$4813$ is the largest number in the expansion among at least the first $100$ numbers.

Note that $\sqbrk {0; 1, 5, 2, 5, 6, 1} = \dfrac {431} {510}$

This gives the estimate $7^{510} \approx 10^{431}$.

$\blacksquare$


Sources

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