# Approximation to Power of 7 by Power of 10

## 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:

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

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

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

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

Then:

 $\displaystyle \size {\log c}$ $=$ $\displaystyle q_k \size {\log 7 - \frac {p_k} {q_k} }$ $\displaystyle$ $<$ $\displaystyle q_k \paren {\frac 1 {q_k q_{k + 1} } }$ Accuracy of Convergents of Continued Fraction $\displaystyle$ $=$ $\displaystyle \frac 1 {q_{k + 1} }$ $\displaystyle$ $=$ $\displaystyle \frac 1 {a_{k + 1} q_k + q_{k - 1} }$ 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.

$4813$ 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 $7^{510} \approx 10^{431}$.

$\blacksquare$