Using awk

[Trevor]

Using the awk tool, write the command that will search through the /etc/passwd
file for an account with the username of "trev1".  If the account name exist, the
command should output the login directory for this account.

Note:  The use of pipes is NOT permitted!!!!

[TM]

awk -F ':' ' $1 == "trev1" { print $6 } ' /etc/passwd

[Trevor]

And we have a winner!!!!  That command syntax will 
absolutely accomplish the task!!!

Great job TM!!!

Another form that would get the job done is:

awk  -F:  ' /trev1/  { print $6 }'  /etc/passwd

[Chetan_Tiwary_]

just to help the learners, here is the explanation of the command and what it does :

First let us analyze the fields in the /etc/passwd file :

same way, as per the question's condition : the login directory of such users is in the 6th field of the /etc/passwd file

trev1:x:1001:1001:Trev User:/home/trev1:/bin/bash

1        2    3      4           5                  6                   7

so, this command when we use it : awk -F ':' ' $1 == "trev1" { print $6 } ' /etc/passwd

awk -F ':' sets the field separator to a colon (:) character. This instructs awk to interpret each line of the input file as a sequence of fields, delimited by ':'.

$1 == "trev1" { print $6 }: The condition $1 == "trev1" checks if the first field of the current line, which represents the username in the /etc/passwd file, matches "trev1".

and then If the username matches "trev1", the command prints the sixth field, which corresponds to the user's home directory in the /etc/passwd file.

and done !!