The question on your screen involves calculating the decibel level $D$ based on a given intensity ratio. The formula provided is:

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

where:

• $I$ is the intensity level,
• $I_0$ is the reference intensity level (often the threshold of hearing).

The problem specifies that $I = 2,511,886 \times I_0$.

Solution Steps

1. Substitute $I = 2,511,886 \times I_0$ into the formula:

   $D = 10 \log \left( \frac{2,511,886 \times I_0}{I_0} \right)$

2. Simplify the fraction:

   $D = 10 \log(2,511,886)$

3. Calculate $\log(2,511,886)$:

   $\log(2,511,886) \approx 6.4$

4. Multiply by 10:

   $D = 10 \times 6.4 = 64 \text{ dB}$

So, the decibel level $D$ is approximately 64 dB.

Would you like further details on this calculation, or do you have any other questions?

Here are five related questions to explore:

1. How does changing the intensity $I$ affect the decibel level $D$?
2. What is the decibel level if $I$ is only 100 times $I_0$?
3. Why do we use logarithms in calculating decibel levels?
4. What would happen if $I = I_0$?
5. How is the reference intensity $I_0$ chosen for different measurements?

Tip: Decibel (dB) scales are logarithmic, meaning each 10 dB increase represents a tenfold increase in intensity.