Come to think of it, this is another context where a logarithmic scale is useful. You couldn’t make a statement like that with a linear scale, because it would depend on how close you were to the source. But you can say it with decibels (or any other logarithmic scale).
Hehe, I remember seeing my dad’s old slide rule in his desk, but never was taught how to use one. I do fondly remember my TI-84 (it’s probably in a box somewhere around here). So now, even I get to feel old.
And here is an example the problem I alluded to earlier. 1000 is an edge case. This is where you need to understand what is going on and manage the rest of the problem.
Since log_{10}(10) = 1. A slide rule doesn’t bring much additional help to the task.
Indeed for any value of x, since x^1 = x we always have log_{x}(x) = 1
In general, however, slide rules do help.
A common slide rule design includes another scale that allows you to read off logarithms. But only in the range 0 to 10. You are expected to manage the rest yourself. (The L scale gives you the log, the D scale the x. The L scale can be above or below depending on design, sometimes you need to use the cursor to line up a reading.) The basic slide rule with only A, B, C and D scales doesn’t help. Which is annoying.
If the question was a 3000 times power level, you would go to 3 on the D scale, and read off the L scale the value 0.47. You have to manage the thousands yourself. Which is easy, you just count the number of zeros. Since 1000 = 10^3, log_{10}(1000) = 3
Since log(xy) = log(x) + log(y) (which is again how a slide rule works)
log_{10}(3 \times 1000) = log_{10}(3) + log_{10}(1000) = 0.47 + 3 = 3.47
And we get 34.7 dB
The edge case of 1000 was answered by simply counting the number of zeros. No additional effort needed.