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 given value of (I) into the formula

Given (I = 10^{32}I_0), and the formula (I(dB)=10\log\left[\frac{I}{I_0}\right]). Substitute (I) into the formula: (I(dB)=10\log\left[\frac{10^{32}I_0}{I_0}\right]).

Step2: Simplify the fraction inside the logarithm

(\frac{10^{32}I_0}{I_0}=10^{32}), so the formula becomes (I(dB)=10\log(10^{32})).

Step3: Use the logarithm property (\log(a^b)=b\log(a))

Since (\log(10^{32}) = 32\log(10)) and (\log(10)=1), then (I(dB)=10\times32).

Step4: Calculate the final result

(10\times32 = 320).

Answer:

320