It's that last bit that makes logarithms really useful for talking about a wide range of different kinds of signals, ranging from acoustic to electrical.
If a given resistor causes you to lose quarter of the power flowing through it, and another causes you to half, you can multiply the two together as (1 - ½)(1 - ¼) = ⅜ = (1 - ⅝) to get that you're losing ⅝ = 62.5% of your power.
On the other hand, you can use logarithms to make that calculation a bit easier to do.
Xandra Granade 🏳️⚧️ /
npub1hz…5lzy3
2024-06-18 02:36:36
in reply to nevent1q…lynw
Author Public Key
npub1hzvzdrx8mnxqy9m4xupkce4krf564qac2yfpupw9w80r88z8zkxsw5lzy3Published at
2024-06-18 02:36:36Event JSON
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