# Thread: Dicibel Scale and Log

1. ## Dicibel Scale and Log

A loud car stereo has a decibel level of 110 dB. How many times as intense as the sound of a loud card stereo is the sound of a rock concert speaker. A rock concert speaker has a decibel level of 150 dB.

This is my work:

B_2 - B_1 = 10log(I_2/I_1)
110-150= 10log(I_2/I_1)
-40/10=10/10log(I_2/I_1)
10^-4=log(I_2/I_1)
0.0001=(I_2/I_1)

Therefore the sound of a loud rock stereo is 0.0001 times as intense as a sound of a rock concert speaker.

HOWEVER, the real question comes here: my answer is 0.0001 times, but the answer in the back says 10000 times. so 10^4 instead of my 10^-4.

I think that the back of the book is wrong though BECAUSE, if a rock concert speaker is more loud than a car stereo, then how can a car stereo be more intense than a rock concert speaker? Am I wrong or right? Explain why please.

2. Originally Posted by skeske1234
A loud car stereo has a decibel level of 110 dB. How many times as intense as the sound of a loud card stereo is the sound of a rock concert speaker. A rock concert speaker has a decibel level of 150 dB.

This is my work:

B_2 - B_1 = 10log(I_2/I_1)
110-150= 10log(I_2/I_1)
-40/10=10log(I_2/I_1)
10^-4=log(I_2/I_1) e^(i*pi): -40/10 = -4 not 10^-4
0.0001=(I_2/I_1)

Therefore the sound of a loud rock stereo is 0.0001 times as intense as a sound of a rock concert speaker.

HOWEVER, the real question comes here: my answer is 0.0001 times, but the answer in the back says 10000 times. so 10^4 instead of my 10^-4.

I think that the back of the book is wrong though BECAUSE, if a rock concert speaker is more loud than a car stereo, then how can a car stereo be more intense than a rock concert speaker? Am I wrong or right? Explain why please.
$B_2 - B_1 = 10log\big{(}\frac{I_2}{I_1}\big{)}$

$\frac{150 - 110}{10} = log\big{(}\frac{I_2}{I_1}\big{)}$

$10^{4} = \frac{I_2}{I_1}$

Therefore the book answer is correct. I suspect you put B_1 and B_2 the other way around without changing the ratio of I_2 and I_1. If you want to do it your way use the log identity:

$log(\frac{a}{b}) = -log(\frac{b}{a})$

3. ty