**The Decibel Scale**

Part A = 10 dB

Part B = 20 dB

Part C = 3,6,9 dB

**Solution Below:**

^{2}changes by a multiplicative factor. The number of decibels increases by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is

Β = 10log(I / I_{0}) dB,

where I_{0} is a reference intensity. For sound waves, I_{0} is taken to be 10^{-12} W/m^{2}. Note that log refers to the logarithm to the base 10.

Part A

_{0})?

Express the sound intensity numerically to the nearest integer.

Just follow the formula:

Β = 10log(I / I_{0})

Β = 10log(10 / 1)

Β = 10log(10)

Β = 10 dB

10 dB

Part B

_{0})?

Express the sound intensity numerically to the nearest integer.

Just follow the formula:

Β = 10log(I / I_{0})

Β = 10log(100 / 1)

Β = 10log(100)

Β = 20 dB

20 dB

Part C

Calculate the change in decibels (ΔΒ_{2}, ΔΒ_{4}, and ΔΒ_{8}) corresponding to f = 2, f = 4, and f = 8.

To find the increase take I_{0} in the general formula to be the initial intensity and then take I to be the factor of increase: I=fI_{0}. Then, log(I / I_{0}) = log(f) and the change in intensity measured in decibels is 10log(f).

The answer is:

3,6,9 dB

3,6,9 dB