Decibel

By what percentage does the sound intensity increase if the sound intensity level increases by 1 dB?

Correct answer:

p =  25.8925 %

Step-by-step explanation:

$$\begin{gathered} L=1\text{ dB } \\ {L}_2=L\mathrm{/}10=1\mathrm{/}10={\displaystyle \frac{1}{10}}=0.1\text{ B } \\ {L}_2=\log (q) \\ q=10^{L_2}=10^{0.1}\doteq 1.2589 \\ p=100\cdot (q-1)=100\cdot (1.2589-1)=25.8925\mathrm{\%} \end{gathered}$$

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