Why is variable interpolation not working for variable assignment (for bash)?

From the "SIMPLE COMMAND EXPANSION" portion of the bash manpage:

SIMPLE COMMAND EXPANSION When a simple command is executed, the shell performs the following expansions, assignments, and redirections, from left to right.

1. The words that the parser has marked as variable assignments (those preceding the command name) and redirections are saved for later processing.

2. The words that are not variable assignments or redirections are expanded. If any words remain after expansion, the first word is taken to be the name of the command and the remaining words are the arguments.

3. Redirections are performed as described above under REDIRECTION.

4. The text after the = in each variable assignment undergoes tilde expansion, parameter expansion, command substitution, arithmetic expansion, and quote removal before being assigned to the variable.

So, it might sound insane, but the "word expansion" (interpolation) of echo $LOG happens in step 2, and the variables actually get properly expanded in step 4... I'm not sure why you're not simply getting a blank line, but wonder if it was from prior test runs?

Both of these work around that, either by separating into multiple "simple commands", or by interpolating $LOG in a sub-shell:

ROOT=/tmp LOG=$ROOT/blah.log ; echo $LOG

ROOT=/tmp LOG=$ROOT/blah.log sh -c 'echo $LOG'

Let me guess: you edited that script on Windows.

Correct? Your scripts have Windows line endings: CR LF instead of just LF which is the unix line ending character. As far as the shell is concerned, that CR is part of the command, so the value of ROOT contains a CR at the end, and when you print it out, the CR moves the cursor back to the beginning of the line.

The fix is not to edit scripts on Windows, or else to make sure your editor uses unix line endings. Be sure to convert your existing script(s).

See also Directories are listed twice and Carriage return issue