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At the dance the DJ is blaring music at 110 decibels. As he fades the music, the sound level decreases about 16% every second.
How long until the sound level falls below 30 decibels? (SHOW WORK!)

Equation:_____________________

Answer: ______________________​

Respuesta :

TheSection

We are given:
Initial sound Intensity: 110 dB
Rate of decreasing sound level: 16% every second


Constructing the equation:
Since the Intensity decreases by 16% of it's current value every second.
Intensity(t) = [tex]110 * (\frac{84}{100})^{t}[/tex]               [where t is the time elapsed]
In this equation, we multiply the initial sound intensity by 84% (since 16% was diminished).
because we are multiplying the same value by 84%, it will keep taking that percentage of the new value, which is what we need here because the music's intensity keeps becoming 84% of the previous value.


Time at which sound level falls to 30 dB:
replacing intensity by 30 in the equation we made.
[tex]30 = 110 * (\frac{84}{100})^{t}[/tex]
[tex]\frac{3}{11} = (\frac{84}{100})^t[/tex]
taking log of both sides.
[tex]log(\frac{3}{11}) = t*log(\frac{84}{100})[/tex]
[tex]\displaystyle \frac{log(\frac{3}{11})}{log(\frac{84}{100})} = t[/tex]

t = 7.452 seconds

semsee45

Answer:

Equation:  [tex]y=110(0.84)^x[/tex]

(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)

Step-by-step explanation:

We can model this as an exponential equation.

General form of an exponential equation: [tex]y=ab^x[/tex]

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:

[tex]\implies y=110(0.84)^x[/tex]

To find how long until the sound level falls below 30 decibels, set y < 30 and solve for x:

[tex]\begin{aligned}110(0.84)^x & < 30\\\\(0.84)^x & < \dfrac{3}{11}\\\\\ln (0.84)^x & < \ln \left(\dfrac{3}{11}\right)\\\\x \ln (0.84) & < \ln \left(\dfrac{3}{11}\right)\\\\x & > \dfrac{\ln \left(\dfrac{3}{11}\right)}{\ln (0.84)}\\\\x & > 7.452008851\end{aligned}[/tex]

Therefore, the sound level falls below 30 decibels after 7.452 seconds (3 dp)

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