When a person wears a hearing aid, the sound intensity level increases by 30 dB. The sound intensity increases by

Options

A. `e^2`
B. `10^3`
C. `30`
D. `10^2`

Explanation

To solve the problem of how much the sound intensity increases when a hearing aid increases the sound intensity level by 30 dB, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Decibel Scale**:
The sound intensity level (B) in decibels (dB) is given by the formula:
\[ B = 10 \log_{10} \left(\frac{I}{I_0}\right) \]
where \(I\) is the sound intensity and \(I_0\) is the reference intensity.

2. **Initial and Final Sound Intensity Levels**:
Let the initial sound intensity level be \(B\) corresponding to intensity \(I_1\). When the hearing aid is used, the sound intensity level increases by 30 dB, so the new level is:
\[ B + 30 = 10 \log_{10} \left(\frac{I_2}{I_0}\right) \]

3. **Setting Up the Equations**:
We can write two equations based on the above information:
- For the initial intensity:
\[ B = 10 \log_{10} \left(\frac{I_1}{I_0}\right) \]
- For the final intensity:
\[ B + 30 = 10 \log_{10} \left(\frac{I_2}{I_0}\right) \]

4. **Subtracting the Two Equations**:
Subtract the first equation from the second:
\[ (B + 30) - B = 10 \log_{10} \left(\frac{I_2}{I_0}\right) - 10 \log_{10} \left(\frac{I_1}{I_0}\right) \]
This simplifies to:
\[ 30 = 10 \log_{10} \left(\frac{I_2}{I_1}\right) \]

5. **Dividing Both Sides by 10**:
Divide both sides by 10:
\[ 3 = \log_{10} \left(\frac{I_2}{I_1}\right) \]

6. **Exponentiating to Solve for the Intensity Ratio**:
To eliminate the logarithm, exponentiate both sides:
\[ 10^3 = \frac{I_2}{I_1} \]
Thus, we find:
\[ \frac{I_2}{I_1} = 1000 \]

7. **Conclusion**:
Therefore, the sound intensity increases by a factor of 1000 when the sound intensity level increases by 30 dB.

### Final Answer:
The sound intensity increases by \(10^3\).