To solve this problem, we are given the function for loudness in decibels:

$D(I) = 10 \log \left(\frac{I}{10^{-12}}\right)$

where:

• $D(I)$ is the decibel level,
• $I$ is the intensity of the sound,
• $10^{-12}$ is the intensity of the quietest sound a human can hear.

We need to find the intensity $I$ of the school fire alarm if it registers as 100 decibels. So, $D(I) = 100$.

Steps to Solve:

1. Set up the equation with $D(I) = 100$:

   $100 = 10 \log \left(\frac{I}{10^{-12}}\right)$

2. Isolate the logarithmic term by dividing both sides by 10:

   $10 = \log \left(\frac{I}{10^{-12}}\right)$

3. Rewrite the equation in exponential form to solve for $I$:

   Recall that if $\log(x) = y$, then $x = 10^y$.

   So,

   $\frac{I}{10^{-12}} = 10^{10}$

4. Solve for $I$ by multiplying both sides by $10^{-12}$:

   $I = 10^{10} \cdot 10^{-12}$

5. Simplify the expression:

   $I = 10^{-2}$

So, the intensity $I$ of the sound of the school fire alarm is $10^{-2}$ watts per square meter (W/m²).

Would you like further details on any of these steps, or do you have any questions?

Additional Questions:

1. How would the intensity change if the decibel level was 110 dB?
2. Can you calculate the decibel level if the intensity was $10^{-3}$ W/m²?
3. What would the formula look like if the reference intensity changed?
4. How does a logarithmic scale help in measuring sound intensity?
5. What is the relationship between decibels and perceived loudness by humans?

Tip:

In logarithmic equations, converting the logarithmic form to exponential form is often a useful technique for solving for unknown values.