An equation for loudness(L) in decibels (dB) of a sound source as a function of its relative intensity (R) is below.
L=10log(R)
a. Solve this equation to find the relative intensity as a function of loudness. That is solve for R.
b. A Led Zeppelin concert reaches a loudness of 85 dB. A more mellow Jack Johnson concert reaches a peak loudness of 77 dB. What is the ratio of the relative intensity of Led Zeppelin to Jack Johnson.
please help me solve this
L=10log(R)
L/10 = log R
10^(L/10) = R
Find R for each of the performers, then form the ratio
dividing by 10 ... L/10 = log(R)
taking antilog ... 10^(L/10) = R
a. To solve for R in the equation L=10log(R), we can follow these steps:
Step 1: Rewrite the equation using the logarithmic property log(R) = L/10.
Step 2: Remove the logarithm by taking the antilogarithm of both sides of the equation. This is done by raising both sides to the power of 10.
10^(log(R)) = 10^(L/10)
Step 3: Simplify the equation. The exponentiation undoes the logarithm, so we are left with R on the left side.
R = 10^(L/10)
Therefore, the equation for the relative intensity (R) as a function of loudness (L) is R = 10^(L/10).
b. Given that Led Zeppelin's loudness is 85 dB (L1) and Jack Johnson's loudness is 77 dB (L2), we can use the equation R = 10^(L/10) to find the ratio of their relative intensities.
First, we need to calculate the relative intensity for each concert:
For Led Zeppelin:
R1 = 10^(85/10) = 10^8.5 ≈ 31622.78
For Jack Johnson:
R2 = 10^(77/10) = 10^7.7 ≈ 1968.98
The ratio of the relative intensities is R1/R2, which equals:
R1/R2 ≈ 31622.78 / 1968.98 ≈ 16.07
Therefore, the ratio of the relative intensity of Led Zeppelin to Jack Johnson is approximately 16.07.