Answer:
You need 4 weights, and the weights are 1; 3; 9 and 27.
Step-by-step explanation:
<em>Since the scale has two plates, we can place weights on either side and also the object so it can be balanced. </em>
This is a key part of the problem, it's not only on the other side of the scale, but on both sides.
Let's do the math now.
If i get two weights, 1 and 3. I can form this combinations.
Object of 1lb = 1
Object of 2lb + 1 weight = 3 weight.
Object of 3lb = 3 weight
Object of 4lb = 1 weight + 3 weight.
So what if i want to add the next weight and that weight to add me the maximum amount of objects. The weight would have to have a difference with the last object plus one. So if i grab 9. 9 minus 4 is 5. And that is a difference with the last object plus 1.
With a weight of 9, now i can add all the integers up to 13lb.
And the next step? Lets add one more. Keeping the last rule, the weight would have to have a difference with the last object plus one. So if i grab 27, 27 minus 13 is 14. And that is a difference witht the last object plus 1.
The sum of all the weights adds up to 40 pounds. And i can balance any integer in the middle.
The formula we are using is p – n = n + 1
Where p is the new weight. and n is the last object we weighted. And the sum of the weights goes up to the last object we can place on the scale, and in this case is 40.
can be converted into the power 
. Therefore we can rewrite the problem as 
and then follow the exponent rules about a power to a power, multiplying 1/2 and 3/4 together.
, which then can be turned into ![\sqrt[8]{10} ^{3x}](https://tex.z-dn.net/?f=%20%5Csqrt%5B8%5D%7B10%7D%20%5E%7B3x%7D)
, making the last option our answer.