the intensity, or loudness, of a sound can be measured in decibels (db), according to the equation…

the intensity, or loudness, of a sound can be measured in decibels (db), according to the equation (i(db)=10logleft\frac{i}{i_{0}}\right), where (i) is the intensity of a given sound and (i_{0}) is the threshold of hearing intensity. what is the intensity, in decibels, (i(db)), when (i = 10^{32}(i_{0}))?\n15\n32\n320\n737

the intensity, or loudness, of a sound can be measured in decibels (db), according to the equation (i(db)=10logleft\frac{i}{i_{0}}\right), where (i) is the intensity of a given sound and (i_{0}) is the threshold of hearing intensity. what is the intensity, in decibels, (i(db)), when (i = 10^{32}(i_{0}))?\n15\n32\n320\n737

Answer

Explanation:

Step1: Substitute the value of (I) into the formula

Given (I(dB)=10\log\left(\frac{I}{I_0}\right)) and (I = 10^{32}I_0), then (I(dB)=10\log\left(\frac{10^{32}I_0}{I_0}\right)). Since (\frac{10^{32}I_0}{I_0}=10^{32}), the formula becomes (I(dB)=10\log(10^{32})).

Step2: Use the logarithm property

We know that (\log(a^b)=b\log(a)), for (a = 10) and (b = 32), (\log(10^{32})=32\log(10)). And since (\log(10) = 1), then (\log(10^{32})=32). So (I(dB)=10\times32).

Step3: Calculate the result

(10\times32 = 320).

Answer:

320