Chapter 3.3, Problem 79E

To determine

To write: logarithm in simpler form and determine the number of decibels of a sound with an intensity of ${10}^{-6}$ watt per square meter.

Answer to Problem 79E

  $\beta =10\left(\log I+12\right)$

  $60\text{ dB}$

Explanation of Solution

Given:

The relationship between the number of decibels $\beta$ and the intensity of a sound I (in watts per square meter) is:

  $\beta =10\log \frac{I}{{10}^{-12}}$

Concept Used:

Quotient property of logarithm states that for a positive real numbers a such that $a\ne 1$ :

  ${\log }_a\left(\frac{u}{v}\right)={\log }_{a}u-{\log }_{a}v$

Power property of logarithm states that for a positive real numbers a such that $a\ne 1$ :

  ${\log }_a{u}^n=n{\log }_{a}u$

Consider,

  $\beta =10\log \frac{I}{{10}^{-12}}$

Using the quotient property,

  $\beta =10\left(\log I-\log {10}^{-12}\right)$

Using power property,

  $\begin{array}{c}\beta =10\left(\log I-\left(-12\right)\log 10\right)\\ =10\left(\log I+12\log 10\right)\end{array}$

Using ${\log }_{b}b=1$ ,

  $\begin{array}{c}\beta =10\left(\log I+12\log 10\right)\\ =10\left(\log I+12{\log }_{10}10\right)\\ =10\left(\log I+12\left(1\right)\right)\\ =10\left(\log I+12\right)\end{array}$

Now substitute $I={10}^{-6}$ in $\beta =10\left(\log I+12\right)$ ,

  $\beta =10\left(\log {10}^{-6}+12\right)$

Using the power property,

  $\beta =10\left(-6\log 10+12\right)$

Using ${\log }_{b}b=1$ ,

  $\begin{array}{c}\beta =10\left(-6\left(1\right)+12\right)\\ =10\left(-6+12\right)\\ =10\left(6\right)\\ =60\text{ dB}\end{array}$

Hence, a sound with an intensity of ${10}^{-6}$ watt per square meter is of 60 dB.