if a sound is increased (decreased) in intensity by a factor of 10, its dB value increases (decreases) by

If a sound's intensity increases by a factor of 10, its decibel (dB) level increases by **10 dB**. Conversely, if the intensity decreases by a factor of 10, its dB level decreases by **10 dB**.

This relationship is based on the logarithmic nature of the decibel scale, where the formula for calculating the sound level in decibels is:

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

Where:
- \( I \) is the intensity of the sound.
- \( I_0 \) is a reference intensity.

Therefore:
- A tenfold increase in intensity corresponds to a 10 dB increase.
- A tenfold decrease in intensity corresponds to a 10 dB decrease.

Would you like more details on how the decibel scale works?

### 5 Related Questions:
1. What is the relationship between sound intensity and perceived loudness?
2. How does the decibel scale differ from a linear scale?
3. What happens to the dB level if sound intensity is increased by a factor of 100?
4. What is the threshold of hearing in decibels?
5. How is sound intensity related to distance from the sound source?

**Tip:** The decibel scale is logarithmic, meaning each 10 dB increase represents a tenfold increase in intensity, but it only seems roughly twice as loud to our ears.

Discover how sound intensity changes affect decibel levels. Learn the logarithmic relationship between sound intensity and decibel values, using the formula dB = 10 * log10(I / I0). Suitable for students in Grades 9-12 studying logarithms and sound intensity.

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