Answer:
Equation:
![y=110(0.84)^x](https://img.qammunity.org/2023/formulas/mathematics/college/x4060cwio2jwvhbbjx40zwomttj5aq6v9b.png)
(where x is the time in seconds, and y is the sound level in decibels)
Answer: The sound level falls below 30 decibels after 7.452 seconds (3 dp)
Explanation:
We can model this as an exponential equation.
General form of an exponential equation:
![y=ab^x](https://img.qammunity.org/2023/formulas/mathematics/high-school/hye5rg1h8wj3ohgdt4j1vpepdhoym0w9ex.png)
where:
- a is the initial value
- b is growth factor
- x is the independent variable
- y is the dependent variable
If b > 1 then it is an increasing function
If 0 < b < 1 then it is a decreasing function
If the sound decreases by 16% each second, then the growth (decay) factor will be 100% - 16% = 84% → 0.84
Given:
- a = 110 decibels
- b = decreases by 16% each second = 0.84
- x = time (in seconds)
- y = sound level (in decibels)
Substituting these values into the equation:
![\implies y=110(0.84)^x](https://img.qammunity.org/2023/formulas/mathematics/college/8dy2yn8izloft02cp6xggpgxizfae9i5y4.png)
To find how long until the sound level falls below 30 decibels, set y < 30 and solve for x:
![\begin{aligned}110(0.84)^x & < 30\\\\(0.84)^x & < (3)/(11)\\\\\ln (0.84)^x & < \ln \left((3)/(11)\right)\\\\x \ln (0.84) & < \ln \left((3)/(11)\right)\\\\x & > (\ln \left((3)/(11)\right))/(\ln (0.84))\\\\x & > 7.452008851\end{aligned}](https://img.qammunity.org/2023/formulas/mathematics/college/apgwxg31gzfi4gbof9kfwcyl2ze5r4yi43.png)
Therefore, the sound level falls below 30 decibels after 7.452 seconds (3 dp)