Question

the loudness level of a sound, d, in decibels, is given by the formula d = 10 log (10^12i), where i is the intensity of the sound, in watts per meter². decibel levels range from 0, a barely audible sound, to 160, a sound resulting in a ruptured eardrum. the sound of a certain animal can be heard 500 miles away, reaching an intensity of 7.4×10^5 watts per meter². determine the decibel level of this sound. at close range, can the sound of this animal rupture the human eardrum? the decibel level of this animals sound is approximately (round to the nearest whole number as needed.)

Explanation:

Step1: Substitute the intensity value.

Given $I = 7.4\times10^{5}$, and the formula $D = 10\log(10^{12}I)$. Substitute $I$ into the formula: $D = 10\log(10^{12}\times7.4\times10^{5})$.

Step2: Simplify the exponent part.

Using the rule of exponents $a^{m}\times a^{n}=a^{m + n}$, we have $10^{12}\times7.4\times10^{5}=7.4\times10^{12 + 5}=7.4\times10^{17}$. So $D = 10\log(7.4\times10^{17})$.

Step3: Use the logarithm property $\log(ab)=\log(a)+\log(b)$.

$D = 10(\log(7.4)+\log(10^{17}))$. Since $\log(10^{17}) = 17$ (because $\log_{10}(10^{x})=x$), then $D = 10(\log(7.4)+17)$.

Step4: Calculate $\log(7.4)$.

$\log(7.4)\approx0.869$. Then $D = 10(0.869 + 17)$.

Step5: Calculate the final value.

$D=10\times17.869 = 178.69\approx179$.

Answer:

179