A Better Estimate of A Big Number!

$\textbf{Solution:}$

Let $a$ be a real number such that $9^a = 689^{1679}$ . Taking $\log$ on both sides yields

$a = \frac{1679 \log 689}{\log 9} \approx 4993.877$ .

Clearly, $4900 < a < 4995$ , for which we can eliminate two choices, $9^{4996}$ and $9^{4997}$ . Now we only compare $9^{4900}$ and $9^{4995}$ .

$\displaystyle \frac{9^{4995} - 9^a}{9^a - 9^{4900}} = \frac{9^{95} - 9^{a-4900}}{9^{a-4900} - 1} > \frac{9^{95} - 9^{a-4900}}{9^{a-4900}} = 9^{4995 - a} -1 > 1$

Now, $|9^{4995} - 689^{1679}| > |689^{1679} - 9^{4900}|$ and so the best estimate is $9^{4900}$ . []

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$\textbf{Alternative Solution}$ :

Let $a$ be a real number such that $9^a = 689^{1679}$ . Taking $\log$ on both sides yields

$a = \frac{1679 \log 689}{\log 9} \approx 4993.877$ .

Clearly, $4900 < a < 4995$ . So $9^{4900} < 9^a < 9^{4995}$ , for which we can eliminate two choices $9^{4996}$ and $9^{4997}$ . Now we consider the arithmetic mean of $9^{4995}$ and $9^{4900}$ :

$\begin{aligned} \displaystyle \log_9 \left(\frac{9^{4900} + 9^{4995}}{2}\right) &=\log_9 \left(\frac{9^{4900}\left(1 + 9^{95}\right)}{2}\right)\\ &= \log_9 {9^{4900}} + \log_9 \left(1 + 9^{95}\right) - \log_9 2\\ & > 4900 + \log_9 9^{95} - \log_9 2\\ &= 4995 - \frac{\log 2}{\log 9}\\ & \approx 4994.68 > a \end{aligned}$

Hence

$\displaystyle 9^{4900} < 9^a < \frac{9^{4900} + 9^{4995}}{2}.$

This means

$|9^{4995} - 689^{1679}| > |689^{1679} - 9^{4900}|$ and so the best estimate is $9^{4900}$ . []