How to get difference between 2 dates in Linux?

I know this is a common question, but I do not understand how the solutions provided work. Everyone advises to change the date to seconds from the epoch and then divide by (24 * 3600) to get the difference in days. Or use

-shell
if [ $first_date -gt $second_date ]

However, what I don't understand is why no one notices the glaring mistake of using seconds from epoch in that they would differ even within a day. Or if you consider 5:30 pm of a specific day and 9:00 am of the next day, the difference between the seconds won't lead to a gap of 24 hours, and thus, they would be erroneously considered to be the same day. My usage scenario is this: I need to purge logs that are older than a given number of days. Thus, my expiration_date, which I am forming after taking the user input of the number of days is:

-shell
expiration_date=$(date -d "-$1 day" + %s)

My file modification dates I am getting by using stat command as follows:

-shell
file_date=$( stat -c %Y $entry )

I need to compare these two dates and purge them if the file modification dates are "less than" the expiration date. Please help me in this regard. EDIT ==== I am really confused as to the use of the mtime argument of the find command. Let's consider that we have the following files:

Nov15_1
Nov15_2
Nov17_1
Nov17_2
Nov18_1
Nov18_2
Nov19_1
Nov19_2

* If I now run the find command using 0 as the mtime argument on Nov 19, it's giving me all the files except the first four.

-shellsession
    $ find /dir/ -type f -mtime 0 -name "Nov*"
    Nov18_1
    Nov18_2
    Nov19_1
    Nov19_2

* And if I run it using +0, it's giving me only the first four files.

-shellsession
    $ find /dir/ -type f -mtime +0 -name "Nov*"
    Nov15_1
    Nov15_2
    Nov17_1
    Nov17_2

If we express this using variables, saying that we want to purge logs from the nth day of the month and further back, and we run the command on the (n+1)th day using mtime as +0, which should translate to "1 day back and further", it's actually taking the files from the (n-1)th day and back, i.e. from "2 days back and further". Does this mean that there is no way to get just the files from 1 day back and further somehow? The argument of 0 is mixing it with today's files as well. EDIT ==== So I incorporated the -daystart option as pointed out by @AdminBee, and results are as per my expectations. Now if I run the command on the (n+1)th day trying to remove the logs of the nth day and earlier, the command would be

-shell
find /dir/ -type f -daystart -mtime +0 -name "Nov*"

Irrespective of whether the file modification timestamp is within the 24-hour period moving back from now, it will now consider the files of the previous day as being behind by 1 day, and will also list all other files previous to them.