Chapter 6.5, Problem 44E

To determine

Tocalculate:The increase in the intensity of television program in decibels.

Answer to Problem 44E

There is a $10$ decibels increase in the intensity of advertisement from that of the television program.

Explanation of Solution

Given information:

The sound of advertisements is $10$ times of a television program.

Formula used:

Logarithmic function

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

Consider thefunction of loudness of advertisements as:

  $L\left(10I\right)$

Consider the function of loudness of television program as:

  $L\left(I\right)$

Subtract the loudness of television program from that of advertisement.

  $L\left(10I\right)-L\left(I\right)$

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

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

Solve the above expression.

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

Hence, there is a $10$ decibels increase in the intensity of advertisement from that of the television program.