Weight Scale Problem

A market vendor sells herbs in whole number amounts from 1 to 40 grams.
Using an old fashioned 2 panbalance and just 4 weights, she can measure any amount needed.
what are the 4 weights?

This problem turned out to be quite the challenge for me.  Although not directly stated in the question it should be noted that weights can be put on either side at any given time.  So for example if you had a 10kg weight and a 15 kg weight you could put the 10 on one side and the 15 on the other, then add 5g of herbs to balance it.  So we know that we need to find 4 weights in which their additive, subtractive, and counter parts contain all integers from 1-40.

I assumed we needed a 1g weight.  I then made a chart with all the numbers on it from 1 - 40 just so I could keep in check which numbers I needed to get.

From there I did a very methodical but time consuming procedure to find out the answer.

Since we already have a 1g weight, we can cross off 1 from our list.  Now we need to satisfy 2.  We can do this by either adding another 1g weight(1+1 = 2), 2g weight, or 3g weight(3-1 =2).  I decided to go with 3 because it was the furthest away from 1.

So now we can cross off 1,2(3-1),3, and 4(3+1) from our list.

Next we want to cross off 5 from our list.  We can do this by adding a 2g weight(2+3=5), a 4g weight (4+1=5), a 5g weight, a 6g weight(6-1=5), a 8g weight(8-3=5), or a 9g weight (9-(3+1)=5).  I decided to go with 9 because it was the furthest away from 3.

So now we can cross off all the way from 1-13 because we have 1,3,9g weights.

5(9-1-3)
6(9-3)
7(9+1-3)
8(9-1)
9
10(9+1)
11(9+3-1)
12(9+3)
13(9+3+1)

Now we need to cross off 14-40.
We know that we need 4 weights that will add up to 40.  So if we already have 1,3,9 what do we need to get 40?  1+3+9+x=40
                          x=27

Now we can cross off all the numbers from 1-40 because we have 1,3,9,27g weights.

This was the only way I could think of approaching this problem and it happened to work out after a decent amount of time spent plugging and chugging.

After seeing the answer 1,3,9,27 it's clear that these are 3^0, 3^1, 3^2, 3^3.  I tried to find a relation/reason for that but I couldn't figure it out.

Teacherly notes

It might not seem obvious at first but being able to recognize that both sides of the pan can be used is clearly essential to this problem so I think this should be stated in the problem.  This problem was really difficult for myself so I don't know if I can really give too much teacherly advice.

An extension could be
A market vendor sells herbs in whole number amounts from 1 to x grams.
Using an old fashioned 2 panbalance
what are the minimum amount of weights needed to find to measure any amount needed?