Question

What is the decibel level of the radio with intensity 10−7 watts per square inch? Use a logarithmic model to solve.

Answer 1

Answer:

$D = 50$

Step-by-step explanation:

Given

$I = 10^{-7}W/m^2$ --- Intensity

Required

Determine the decibel level (D)

This is calculated as:

$D = 10 * log(\frac{I}{I_n})$

Where:

$I_n =$ The threshold intensity

$I_n =1 * 10^{-12}W/m^2$

So, we have:

$D = 10 * log(\frac{I}{I_n})$

This gives:

$D = 10 * log(\frac{10^{-7}W/m^2}{1 * 10^{-12}W/m^2})$

$D = 10 * log(\frac{10^{-7}}{10^{-12}})$

Apply law of indices

$D = 10 * log(10^{-7--12})$

$D = 10 * log(10^{5})$

Apply law of logarithm

$loga^b = b\ log(a)$

So, we have:

$D = 10 * 5 * log(10)$

$log(10) =1$

So:

$D = 10 * 5 *1$

$D = 50$

Hence, the decibel level is 50