The image contains the following problem:

Given the formula: $L_2 = L_1 + 20 \log \left( \frac{r_1}{r_2} \right)$ where:

• $L_1$ is the sound level (in dB) at distance $r_1$,
• $L_2$ is the sound level (in dB) at distance $r_2$,
• $r_2 > r_1$.

You are asked to determine at what distance the volume is reduced by 6 dB, meaning $L_2 = L_1 - 6$.

Step-by-Step Solution:

1. Set up the equation for the reduction in sound level: $L_2 = L_1 - 6$ Substituting into the original equation: $L_1 - 6 = L_1 + 20 \log \left( \frac{r_1}{r_2} \right)$

2. Simplify the equation: $-6 = 20 \log \left( \frac{r_1}{r_2} \right)$

3. Solve for $\frac{r_1}{r_2}$ by isolating the logarithmic term: $\log \left( \frac{r_1}{r_2} \right) = \frac{-6}{20} = -0.3$

4. Exponentiate both sides to remove the logarithm: $\frac{r_1}{r_2} = 10^{-0.3}$ Using a calculator: $10^{-0.3} \approx 0.5012$

5. Solve for $r_2$: $r_2 = \frac{r_1}{0.5012} \approx 1.996 \cdot r_1$

Thus, the distance $r_2$ is approximately 2 times the original distance $r_1$ when the volume is reduced by 6 dB.

Would you like further details or have any questions?

Here are some related questions:

1. How does the inverse square law relate to sound propagation?
2. What happens to the sound intensity if the distance is doubled?
3. How does the logarithmic scale apply to sound measurements?
4. What is the difference between sound intensity and sound pressure level?
5. How does frequency affect sound levels at different distances?

Tip: Sound decreases logarithmically with distance, so small changes in dB can represent large changes in distance.