This happened more than a week ago but I copied the sys.log and auth.log . In auth.log I found this entry:

Mar 26 09:51:10 My-PC pkexec: pam_unix(polkit-1:session): session opened for user root by (uid=1000)

Mar 26 09:51:10 My-PC pkexec: pam_systemd(polkit-1:session): Cannot create session: Already running in a session

Mar 26 09:51:10 My-PC pkexec[12500]: computer: Executing command [USER=root] [TTY=unknown] [CWD=/home/computer] [COMMAND=/usr/lib/unity-settings-daemon/usd-backlight-helper --set-brightness 158]

Mar 26 10:17:01 My-PC CRON[12756]: pam_unix(cron:session): session opened for user root by (uid=0)

I searched for pkexec[12500] and I found out that pkexec allows an authorized user to execute PROGRAM as another user.. What does this mean? Could this command be used by a hacker?

Later at 1 PM the computer shut off by itself. I found this in auth.log:

Mar 26 13:00:14 My-PC systemd-logind[1265]: System is powering down. Mar 26 13:00:14 My-PC polkitd(authority=local): Unregistered Authentication Agent for unix-session:c2 (system bus name :1.75, object path /org/gnome/PolicyKit1/AuthenticationAgent, locale en_US.UTF-8) (disconnected from bus)

What could this mean?

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