Question

The intensity, or loudness, of a sound can be measured in decibels (dB), according to the equation I(dB)=10log[1/10], where I &/ the intensity of a given sound and I0 is the threshold of a hearing intensity. What is the intensity, in decibles, [I(dB)], when I=10^8(I0)? Round to the nearest whole number. (Refer to photo for equations) A.8 B.9 C.19 D.80

Answer 1

For this case we have the following equation:
 $I(dB)= 10log(\frac{I}{I0})$
 Where,
 I: the intensity of a given sound
 I0:  the threshold of a hearing intensity
 Substituting values in the given equation we have:
 $I(dB)= 10log(\frac{10^8I0}{I0})$
 Rewriting the equation we have:
 $I(dB)= 10log(10^8)$
 Then, by properties of logarithm we have:
 $I(dB)= 10(8)$
 Finally we have:
 $I(dB)=80$
 Answer:
 The intensity, in decibles, [I(dB)], when I=10^8(I0) is:
 D.80

Answer 2

Answer:

D) 80

Step-by-step explanation: