Solve for the indicated value, and graph the situation showing the solution point. The formula for measuring sound intensity in decibels DD is defined by the equation D=10log(II0)D=10logII0 using the common (base 10) logarithm where II is the intensity of the sound in watts per square meter and I0=10−12I0=10−12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of 8.1⋅1028.1⋅102 watts per square meter?
 
Round your answer to three decimal places.
 
The jet plane emits  

 decibels at 8.1⋅1028.1⋅102 watts per square meter


To solve this problem, we'll use the formula for measuring sound intensity in decibels:

\[
D = 10 \log \left( \frac{I}{I_0} \right)
\]

where:
- \( D \) is the sound intensity in decibels,
- \( I \) is the intensity of the sound in watts per square meter,
- \( I_0 = 10^{-12} \) is the reference intensity, the lowest level of sound that the average person can hear.

We are given:
- \( I = 8.1 \times 10^{2} \) watts per square meter.

Let's plug in the values and solve for \( D \).

\[
D = 10 \log \left( \frac{8.1 \times 10^{2}}{10^{-12}} \right)
\]

Simplify the fraction inside the logarithm:

\[
\frac{8.1 \times 10^{2}}{10^{-12}} = 8.1 \times 10^{2 + 12} = 8.1 \times 10^{14}
\]

Now, compute the logarithm:

\[
D = 10 \log \left( 8.1 \times 10^{14} \right)
\]

Using the properties of logarithms, we can separate this into two parts:

\[
D = 10 \left( \log (8.1) + \log \left(10^{14}\right) \right)
\]

Since \(\log \left( 10^{14} \right) = 14\), we have:

\[
D = 10 \left( \log (8.1) + 14 \right)
\]

Next, we calculate \(\log (8.1)\).

\[
\log (8.1) \approx 0.908
\]

Substitute this value into the equation:

\[
D = 10 \left( 0.908 + 14 \right)
\]
\[
D = 10 \times 14.908 = 149.08
\]

Rounding to three decimal places:

\[
D \approx 149.080 \text{ decibels}
\]

So, the jet plane emits approximately **149.080 decibels** at \( 8.1 \times 10^{2} \) watts per square meter.

### Graphing the Situation
To graph this, we can plot the sound intensity \( I \) on the x-axis (in logarithmic scale due to the wide range of values) and the decibel level \( D \) on the y-axis. The solution point will be marked where \( I = 8.1 \times 10^{2} \) watts per square meter and \( D = 149.080 \) decibels.

If you need a detailed graph, I can provide that as well. Would you like to see the graph and any further details?

### Follow-up Questions
1. What is the decibel level for a sound intensity of \( 1 \times 10^{-6} \) watts per square meter?
2. How does changing the intensity \( I \) by a factor of 10 affect the decibel level \( D \)?
3. What is the decibel level at the threshold of hearing (\( I = 10^{-12} \) watts per square meter)?
4. What is the decibel level of a normal conversation if its intensity is \( 1 \times 10^{-6} \) watts per square meter?
5. How does the decibel level change if the intensity is doubled?

### Tip
Remember that the decibel scale is logarithmic, not linear. This means a small increase in decibel level represents a large increase in actual sound intensity!

Solve for the indicated value, and graph the situation showing the solution point. The formula for measuring sound intensity in decibels is defined by the equation D = 10 log(I / I0) using the common (base 10) logarithm where I is the intensity of the sound in watts per square meter and I0 = 10^-12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of 8.1⋅10^2 watts per square meter? Round your answer to three decimal places.

Solve for the decibel level of a jet plane emitting sound at an intensity of 8.1⋅10^2 watts per square meter. Using the formula D = 10 log(I / I0), this detailed solution explains each step and includes a graph of the situation. Suitable for students in Grades 10-12.

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