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 $i(db)=10\\log\\left\\frac{i}{i_0}\\right$, 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 $i(db)=10\\log\\left\\frac{i}{i_0}\\right$, 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 $I = 10^{32}I_0$ into the formula

$I(dB)=10\log\left(\frac{10^{32}I_0}{I_0}\right)$

Step2: Simplify the fraction inside the logarithm

Since $\frac{10^{32}I_0}{I_0}=10^{32}$, then $I(dB)=10\log(10^{32})$

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

Here $a = 10$, $b = 32$, so $\log(10^{32})=32\log(10)$. And since $\log(10) = 1$, we have $\log(10^{32})=32$

Step4: Calculate the final value

$I(dB)=10\times32 = 320$

Answer:

C. 320