The formula for the loudness $L(x)$, measured in decibels (dB), is given by:

$L(x) = 10 \log \left( \frac{x}{I_0} \right)$

where $I_0 = 10^{-12} \, \text{watt/m}^2$ is the least intense sound a human ear can detect, and $x$ is the intensity of the sound in watts per square meter.

We are given:

• $x = 10^{-9.4} \, \text{watt/m}^2$
• $I_0 = 10^{-12} \, \text{watt/m}^2$

Substitute these values into the formula for $L(x)$:

$L(x) = 10 \log \left( \frac{10^{-9.4}}{10^{-12}} \right)$

Simplify the argument inside the logarithm:

$\frac{10^{-9.4}}{10^{-12}} = 10^{-9.4 + 12} = 10^{2.6}$

Now the formula becomes:

$L(x) = 10 \log(10^{2.6})$

Since $\log(10^{2.6}) = 2.6$, we have:

$L(x) = 10 \times 2.6 = 26 \, \text{decibels}$

Thus, the loudness of the sound is 26 decibels.

Would you like further clarification, or do you have any questions?

Here are 5 related questions for practice:

1. What is the loudness of a sound with intensity $10^{-8} \, \text{watt/m}^2$?
2. How does doubling the intensity of sound affect the decibel level?
3. If a sound has a loudness of 50 decibels, what is its intensity in watts per square meter?
4. What is the logarithmic relationship between sound intensity and loudness?
5. How does a 10 dB increase affect perceived loudness by the human ear?

Tip: Remember, a 10 dB increase roughly doubles the perceived loudness of a sound to the human ear!