OK, at the end of the previous post, I had decided that if there was a 1 watt light bulb inside a round balloon that had a radius of 1 meter, then the surface area of the inside of the balloon was about 12 square meters, and each square meter would get 1/12 of the power from the bulb, or 1/12th of a watt.
This all came about because of this formula I found on the internet:
Surface Area of a Sphere = 4 pi r 2
I had rounded pi off to 3 to make it easy to figure. And having r be 1 meter was nice too. That’s my kind of math.
But, what if the balloon were bigger; what if the radius were 2 meters, not 1 meter? Well, in that case, we would have to figure out a new answer. The only thing that is different is r, it changes from 1m to 2m. In the equation, r is squared, so instead of dealing wih (1m) 2 now we are dealing with (2m) 2 . Well, the superscript 2 after the parantheses means to multiply what is inside by itself. So that is just 2 meters * 2 meters, or 4 square meters.
We could go ahead and figure out the rest of the equation, but why bother. We know the rest did not change, and it is obvious that this part went from 1 to 4, or got 4 times bigger. So there will be 4 times more square meters than before. So each square meter will get only 1/4th of what it got before. That is what is important. When we double the distance from the light bulb to the balloon surface, the power for each square meter at the surface got smaller by 4 times!
And if we double the radius of the balloon again, to 4 meters, guess what? The power at the surface will decrease by another factor of 4, so now it will be only 1/16 of what it was at 1 meter.
Just to be complete, we need to double the radius one more time, to 8 meters. Now, the power that hits 1 square meter on the inside of the balloon will decrease again, by the same factor of 4. The power hitting the inside surface always drops by 4 when we double the distance. So now, for a radius of 8 meters, we have a drop in power of 4*4*4, or a decrease by a factor of 64. Or 1/64th of the power at each square meter.
OK, we did the increase by 2, and if we figure out what happens if we increase the radius by a multiple of10, then we could have some nice tools at our disposal. So let’s plug r=10 meters into the formula:
Surface Area of our Sphere = 4 pi r 2 = 4*3*(10m) 2 = 4*3*100m 2 .
Now we can see that the radius squared part changed from 1 square meter to 100 square meters. And 4*3 is still 4*3, so lets ignore it.
So we learned that by increasing the radius by 10, we decreased the power hitting any 1 square meter by 100.
Now we have enough information we could make a chart, just remember the chart. Then we can try and avoid remembering the formula, and pi, and all that. The chart might look like this:
distance multiple: power decreases by:
2 1/4th
4 1/16th
8 1/64th
10 1/100th
Now, we can consider almost any distance change just by using these multiples in sequence. For example, if we increase the distance by 40, that is just 4 * 10, so the decrease is just (1/16th) * (1/100), or 1/1600.
I am working my way towards radio here, and these numbers can be clumsy to use. In this case, when I started with a radius of 1, it is obvious the decrease in power is just the square of the new radius, and inverted if you want it as a fraction. But in real life you don’t always get to start at one. The table of change in power with distance change always applies though, regardless of where we start. So except for the darned fractions, the need to multiply and divide, and the hint we see that the numbers might be to bit to do in our head, we have a system. It just needs refined. And to refine it, we need to change to a different measuring system that is easier to work with. So we will introduce logarithms so we can add and subract instead of multiply and divide, and it will help keep the numbers within a range of -100 to +100 typically. Most of us can add and subract in that range! 
