Question

a wind turbine uses the power of wind to generate electricity. the blades of the turbine make a noise that can be heard at a distance from the turbine. at a distance of ( d = 0 ) meters from the turbine, the noise level is 105 decibels. at a distance of ( d = 100 ) meters from the turbine, the noise level is 49 decibels. the noise level can be modeled by the function ( s ) given by ( s(d)=acdot b^{d} ), where ( s(d) ) is the noise level, in decibels, at a distance of ( d ) meters from the turbine. use the model to find the value of ( m ) such that the noise level is 20 decibels at a distance of meters from the turbine.

Explanation:

Step1: Find the value of \(a\)

When \(d = 0\), \(S(0)=a\cdot b^{0}\). Since \(b^{0}=1\) for \(b
eq0\), and \(S(0) = 105\), we have \(a = 105\).

Step2: Find the value of \(b\)

When \(a = 105\) and \(d = 100\), \(S(100)=105\cdot b^{100}\). Given \(S(100)=49\), then \(105b^{100}=49\). So \(b^{100}=\frac{49}{105}=\frac{7}{15}\), and \(b =(\frac{7}{15})^{\frac{1}{100}}\).

Step3: Solve for \(m\)

We know \(S(m)=20\), and \(S(d)=105\cdot b^{d}\). Substitute \(a = 105\) and \(b =(\frac{7}{15})^{\frac{1}{100}}\) into the equation: \(20 = 105\cdot(\frac{7}{15})^{\frac{m}{100}}\).
First, \((\frac{7}{15})^{\frac{m}{100}}=\frac{20}{105}=\frac{4}{21}\).
Take the logarithm of both sides. Let's use the natural logarithm: \(\frac{m}{100}\ln(\frac{7}{15})=\ln(\frac{4}{21})\).
Then \(m = 100\times\frac{\ln(\frac{4}{21})}{\ln(\frac{7}{15})}\).
Using a calculator, \(\ln(\frac{4}{21})\approx\ln(4)-\ln(21)\approx1.386 - 3.045=- 1.659\), \(\ln(\frac{7}{15})\approx\ln(7)-\ln(15)\approx1.946-2.708=- 0.762\).
\(m = 100\times\frac{-1.659}{-0.762}\approx218\).

Answer:

\(m\approx218\)