soft question - Order of magnitude

When finding out the order of magnitude of quantities, as said in my textbook, we compare the numerical part with $3.2$ (approximately $\sqrt{10}$ or rounded off version of $3.162$)

Thus, $9.12 \times 10^5$ has an order of magnitude of $6$ since $9.12$ is $>=$ $3.2$

But what if we had $3.17 \times 10^5$. According to my textbook, its order of magnitude would be $5$, as $3.17$ is less than $3.2$. But, as we have to compare with $\sqrt{10}$, which is $3.162$, the order of magnitude should be $6$, as $3.17$ is greater than $3.162$. This is my question.

Or to rephrase, when we are rounding off $\sqrt{10}$ to $3.2$, why aren't we rounding off the numerical part $3.17$ ?

I hope I made myself clear. I have looked at this, this, and this, but that doesn't satisfy me.

Answer

I think you are attaching excessive significance to your order of magnitude estimate. If I say something is $10^5$ to within an order of magnitude I mean that the $\log_{10}$ of that quantity is nearer $5$ than $4$ or $6$. If some quantity has a $\log_{10}$ of about $5.5$ does it really matter whether our order of magnitude estimate is $10^5$ or $10^6$?

Incidentally, if you wondered why we use $\sqrt{10}$ as the dividing line it's because $\log_{10}(\sqrt{10}) = 0.5$. So for example the $\log_{10}$ of something slightly less than $\sqrt(10)$ is nearer $0$ than $1$, while the $\log_{10}$ of something slightly greater than $\sqrt(10)$ is nearer $1$ than $0$.