Converting exponents to scientific notation

Question

I have to solve or estimate the answer to an equation that is as follows:

$$P_\text{blocks} = \frac{398 \cdot 19^{65}}{\prod^{66}_{i=0} 78804 - i}$$

It doesn't take long to realize that this is an extremely small number. So small, in fact, that I can't find a calculator that doesn't return $\frac{1}{\infty}$.

I can estimate the numerator and denominator to make it a little more simple to understand:

$$P_\text{blocks} = \frac{5\cdot10^{85}}{8^{363}}$$

Is there a way to write this in scientific notation, or is it too large to deal with?

Answer 1

Wolfram Alpha gives your approximation as around $1.9196\times10^{-153}$.

http://bit.ly/Xbgkks

However, it gives the original expression to be around $4.5954\times10^{-243}$

http://bit.ly/XbgWqg

So I think something's gone wrong here because you seem to be off by factor of $4\times10^{89}$.

Answer 2

This is not the simplest form technology can reach. It is trivial for an experienced programmer to solve this problem directly (by creating an unlimited-size data type). I'm not going to do it unless you really, really need it (just ask).

You could estimate $8^{264}$ fairly simply if you don't need an exact solution, though. The easiest way is to note that $2^{10}$ is roughly equal to $10^3$, so this suggests every $8^{10}$ adds a little more than nine digits. So you're probably looking at around $10^{79}$ as the denominator (I rounded up), but maybe a few factors larger due to the shift of $1024/1000$ (which you could further use to refine the estimate).

EDIT: Based on Wolfram Alpha there may be something wrong with your initial estimate in terms of factors of $10$ over factors of $8$. Your initial "simplification" estimate is certainly greater than unity even if my math powers estimate is way off.

EDIT EDIT: You might be surprised to learn that MS calculator (the one built into Windows) is remarkably accurate as an engineering calculator. The denominator is roughly $1.14\times10^{328}$, and the whole shebang is roughly $4.60\times10^{-243}$.

For any conceivable applied use, this number is utterly ridiculous. A better estimate: $0$.

Answer 3

The traditional way of doing this would be with logarithms which (base $10$) would give a calculation like $$2.599883072+65\times 1.278753601 -(4.896548262+\cdots + 4.896184379)$$ which is about $-242.337678$ or $\overline{243}.662322$ depending on whether you are using a calculator or tables and taking the anti-logarithm gives about $4.5954 \times 10^{-243}$.