Find the intensity in decibels [tex]$[I( dB )]$[/tex] for each value of [tex]$I$[/tex].

Normal conversation:
[tex]\[
\begin{array}{l}
I = 10^6 I_0 \\
I( dB ) = \square
\end{array}
\][/tex]

Power saw at 3 feet:
[tex]\[
\begin{array}{l}
I = 10^{13} I_0 \\
I( dB ) = \square
\end{array}
\][/tex]

Jet engine at 100 feet:
[tex]\[
\begin{array}{l}
I = 10^{18} I_0 \\
I( dB ) = \square
\end{array}
\][/tex]

[tex]\[\square\][/tex]



Answer :

To find the intensity in decibels for each value of [tex]\( I \)[/tex], we use the formula for calculating the intensity level in decibels:

[tex]\[ I(\text{dB}) = 10 \cdot \log_{10} \left( \frac{I}{I_0} \right) \][/tex]

where [tex]\( I \)[/tex] is the intensity of the sound, and [tex]\( I_0 \)[/tex] is the reference intensity.

Let's calculate each value step by step:

### Normal Conversation:
Given:
[tex]\[ I = 10^6 I_0 \][/tex]

Using the formula:
[tex]\[ I(\text{dB}) = 10 \cdot \log_{10}(10^6) = 10 \cdot 6 = 60 \text{ dB} \][/tex]

So, for a normal conversation, the intensity in decibels is:
[tex]\[ I(\text{dB}) = 60 \text{ dB} \][/tex]

### Power Saw at 3 feet:
Given:
[tex]\[ I = 10^{11} I_0 \][/tex]

Using the formula:
[tex]\[ I(\text{dB}) = 10 \cdot \log_{10}(10^{11}) = 10 \cdot 11 = 110 \text{ dB} \][/tex]

So, for a power saw at 3 feet, the intensity in decibels is:
[tex]\[ I(\text{dB}) = 110 \text{ dB} \][/tex]

### Jet Engine at 100 feet:
Given:
[tex]\[ I = 10^{18} I_0 \][/tex]

Using the formula:
[tex]\[ I(\text{dB}) = 10 \cdot \log_{10}(10^{18}) = 10 \cdot 18 = 180 \text{ dB} \][/tex]

So, for a jet engine at 100 feet, the intensity in decibels is:
[tex]\[ I(\text{dB}) = 180 \text{ dB} \][/tex]

Therefore, we have the following results:

- Normal conversation: [tex]\( I(\text{dB}) = 60 \text{ dB} \)[/tex]
- Power saw at 3 feet: [tex]\( I(\text{dB}) = 110 \text{ dB} \)[/tex]
- Jet engine at 100 feet: [tex]\( I(\text{dB}) = 180 \text{ dB} \)[/tex]

These values represent the intensity levels in decibels for each corresponding sound intensity.