# Two sounds have measured intensities of I1 = 110 W/m2 and I2 = 330.00 W/m2. By how many decibels is the level of sound 1 lower than that of sound 2? Answer in dB.

## db = 10*Log(I1/I2) = 10*Log(110/330) =

-4.77

I1 = 4.77 db below I2.

## To calculate the difference in decibels between two sound levels, we can use the formula:

Difference in decibels (dB) = 10 * log10 (I2 / I1)

where I1 and I2 represent the intensities of sound 1 and sound 2, respectively.

Given the intensities provided:

I1 = 110 W/m2

I2 = 330 W/m2

Substituting these values into the formula, we get:

Difference in decibels (dB) = 10 * log10 (330 / 110)

To calculate this, follow these steps:

1. Divide the intensity of sound 2 by the intensity of sound 1: (330 / 110) = 3.

2. Take the logarithm base 10 of the result: log10 (3) ≈ 0.477.

3. Multiply the logarithm by 10: 0.477 * 10 = 4.77 dB (rounded to two decimal places).

Therefore, the level of sound 1 is approximately 4.77 dB lower than that of sound 2.