# How do "decibels" work (in an easily, layman's understandable way)? **URL:** \ **Category:** Factual Questions\ **Tags:** science-math\ **Created:** [September 13, 2026, 9:03am UTC](https://boards.straightdope.com/t/how-do-decibels-work-in-an-easily-laymans-understandable-way/1033031 "2026-09-13T09:03:03Z")\ **Posts on this page:** 3\ **Page:** 5
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [October 3, 2026, 11:10pm UTC](https://boards.straightdope.com/t/how-do-decibels-work-in-an-easily-laymans-understandable-way/1033031/81 "2026-10-03T23:10:50Z")
> [@Francis\_Vaughan](#): > > So a thousand people is 30dB louder than one person. 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). ---
**Author:** ![scabpicker](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/scabpicker/32/8268_2.png) [@scabpicker](https://boards.straightdope.com/u/scabpicker)\ **Post date:** [October 4, 2026, 4:59am UTC](https://boards.straightdope.com/t/how-do-decibels-work-in-an-easily-laymans-understandable-way/1033031/82 "2026-10-04T04:59:33Z")
> [@Chronos](#): > > Pretty much every high schooler any more has a TI-84 graphing calculator, and all tests allow use of it. But that’s (finally) going obsolete now, too, because of Desmos, a free online graphing calculator which is much better, which is also allowed on almost all tests. 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. ---
**Author:** ![Francis\_Vaughan](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/francis_vaughan/32/3093_2.png) [@Francis\_Vaughan](https://boards.straightdope.com/u/Francis_Vaughan)\ **Post date:** [October 4, 2026, 5:20am UTC](https://boards.straightdope.com/t/how-do-decibels-work-in-an-easily-laymans-understandable-way/1033031/83 "2026-10-04T05:20:05Z")
> [@Reply](#): > > (And how we do calculate it with a slide rule?) 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. [Previous page](https://boards.straightdope.com/t/how-do-decibels-work-in-an-easily-laymans-understandable-way/1033031.md?page=4)