Puzzle: The scales problem

The first thing that we need to know for solving this puzzle is weights and herbs could be placed on the same side of the scale. 

When we measure 1g herbs, we need a 1g weight. 

When we measure 2g herbs, let us try to find a weight that could be as heavy as possible. Don't forget we have a 1g weight. We place 1g weight on the same side with herbs. So we can put a weight with a maximum of 3g on the other side of the scale.

Now we have two weights, 1g and 3g. Keep going, do the combination.

When we measure 5g herbs, 3g weight and 1g weight are not enough to balance the scale. We need a new weight and hope it could be as heavy as possible. Using the same logic we used when measuring 2g herbs, we can have a weight with a maximum of 9g. 

Keep going, do the combination. 

By the same logic, when we run out of weights, the new weight will be  27 = 14 + 9 + 3 + 1.

The values of the four weights are 1g, 3g, 9g, and 27g. I checked these 4 numbers for herbs weights from 15 to 39 on my rough paper, and they all meet the requirement.

I think 1g, 3g, 9g, and 27g is the only solution because the total weight of these four weights is just 40g.

Extension:

Since in the above puzzle, the solution can be written as 3^0, 3^1,3^2, 3^3, I would like to ask my students to try 2^0, 2^1,2^2, 2^3,2^4, and 4^0, 4^1,4^2.

I want them to use these weights to weigh out the amounts of herbs from 1 to some number. They need to find out what this " some number" would be.