When an intensity decreases tenfold, what is the relative change in decibels?

When discussing changes in intensity in the context of decibels, it's crucial to understand the logarithmic nature of the decibel scale. The decibel level is defined as:

[ dB = 10 \log_{10} \left( \frac{I}{I_0} \right) ]

where ( I ) is the intensity being measured and ( I_0 ) is the reference intensity.

When the intensity decreases by a factor of ten, you can plug this change into the formula. A decrease in intensity by a factor of 10 means that ( I = \frac{I_0}{10} ).

Substituting into the formula gives:

[ dB = 10 \log_{10} \left( \frac{I_0/10}{I_0} \right) = 10 \log_{10} \left( \frac{1}{10} \right) ]

This simplifies to:

[ dB = 10 \log_{10} (1) - 10 \log_{10} (10) = 10 \times 0 - 10 \times 1 = -10 , dB. ]

Thus, a ten