Chapter 6.5, Problem 43E

To determine

Tocalculate:The increase in the intensity of sound made by blue whale in decibels.

Answer to Problem 43E

The required increase in the intensity of blue whale from that of the human ear is $60$ decibels.

Explanation of Solution

Given information:

The sound of blue whale is $1000000$ times of a human year.

Formula used:

Logarithmic function

  $L\left(I\right)=10\log \frac{I}{I_o}$

Consider thefunction of loudness of blue whale as:

  $L\left(1000000I\right)$

Consider the function of hearing capacity of human ear as:

  $L\left(I\right)$

Subtract the loudness of television program from that of advertisement.

  $L\left(1000000I\right)-L\left(I\right)$

Use the formula of logarithmic function to solve the above expression, $L\left(I\right)=10\log \frac{I}{I_o}$ .

  $\begin{array}{c}L\left(1000000I\right)-L\left(I\right)\\ =10\log \frac{1000000I}{I_o}-10\log \frac{I}{I_o}\\ =10\left(\log \frac{1000000I}{I_o}-\log \frac{I}{I_o}\right)\\ =10\log 1000000\end{array}$

Use a calculator to find the solution of the expression.

  $\begin{array}{c}10\log 1000000\\ =60\end{array}$

Hence, there is a $60$ decibels increase in the intensity of blue whale from that of the human ear.