An audio engineer is working on sound attenuation for a concert hall. The sound intensity of a particular instrument decreases exponentially over distance. Initially, at 1 meter from the source, the sound intensity is at 100 units. Every additional meter away from the source, the sound intensity decreases to 70% of its intensity at the previous meter. The recursive formula for this situation is I(d)=I(d−1)⋅0.70

Question

, where I(d)
is the sound intensity at d meters from the source, and I(1)=100
. Determine the distance from the source, in meters, at which the sound intensity will be reduced to 5 units. (1 point)

Answer 1

To find the distance from the source at which the sound intensity will be reduced to 5 units, we start with the recursive formula provided:

\[ I(d) = I(d - 1) \cdot 0.70 \]

with the initial condition:

\[ I(1) = 100 \]

We can express \(I(d)\) in terms of \(d\):

1. For 1 meter: \[ I(1) = 100 \]

2. For 2 meters: \[ I(2) = I(1) \cdot 0.70 = 100 \cdot 0.70 = 70 \]

3. For 3 meters: \[ I(3) = I(2) \cdot 0.70 = 70 \cdot 0.70 = 49 \]

4. For 4 meters: \[ I(4) = I(3) \cdot 0.70 = 49 \cdot 0.70 \approx 34.3 \]

5. For 5 meters: \[ I(5) = I(4) \cdot 0.70 \approx 34.3 \cdot 0.70 \approx 24.01 \]

6. For 6 meters: \[ I(6) = I(5) \cdot 0.70 \approx 24.01 \cdot 0.70 \approx 16.807 \]

7. For 7 meters: \[ I(7) = I(6) \cdot 0.70 \approx 16.807 \cdot 0.70 \approx 11.765 \]

8. For 8 meters: \[ I(8) = I(7) \cdot 0.70 \approx 11.765 \cdot 0.70 \approx 8.2355 \]

9. For 9 meters: \[ I(9) = I(8) \cdot 0.70 \approx 8.2355 \cdot 0.70 \approx 5.76485 \]

10. For 10 meters: \[ I(10) = I(9) \cdot 0.70 \approx 5.76485 \cdot 0.70 \approx 4.035395 \]

Now we can summarize:

• \(I(8) \approx 8.2355\)
• \(I(9) \approx 5.76485\)
• \(I(10) \approx 4.035395\)

Since \(I(9) \approx 5.76485\) and \(I(10) \approx 4.035395\), we find that the sound intensity decreases to 5 units between 9 and 10 meters. To find the exact point where it reaches 5 units more precisely, we can derive a formula for \(I(d)\):

Since \(I(d)\) decreases exponentially, we can express it as:

\[ I(d) = 100 \cdot (0.70)^{d - 1} \]

We need to determine \(d\) where \(I(d) = 5\):

\[ 100 \cdot (0.70)^{d - 1} = 5 \]

Dividing both sides by 100:

\[ (0.70)^{d - 1} = 0.05 \]

To solve for \(d - 1\), take the logarithm of both sides:

\[ \log((0.70)^{d - 1}) = \log(0.05) \]

Using the logarithmic identity \(\log(a^b) = b \cdot \log(a)\):

\[ (d - 1) \cdot \log(0.70) = \log(0.05) \]

Now, solve for \(d - 1\):

\[ d - 1 = \frac{\log(0.05)}{\log(0.70)} \]

Now, calculating the values using a calculator:

\[ \log(0.05) \approx -1.3010 \] \[ \log(0.70) \approx -0.155 \]

Now substituting these values in:

\[ d - 1 = \frac{-1.3010}{-0.155} \approx 8.387 \]

Thus:

\[ d \approx 9.387 \]

So, the distance from the source at which the sound intensity is reduced to 5 units is approximately:

\[ \boxed{9.39} \text{ meters} \]