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Z definicji poziomu natężenia dźwięku: L = 10·log(I/Io) , więc:
∆L = 10·log(I2/Io) - 10·log(I1/Io) = 10·[log(I2/Io) - log(I1/Io)]
Po zastosowaniu wzoru na odejmowanie logarytmów:
∆L = 10·log[(I2/Io)/(I1/Io)] = 10·log(I2/I1) = 10·log2≈ 10·0.3 = 3 dB