Question

The loudness L of sound in decibels is given by L=10log(I/I0) where I is the intensity of the sound, and I0 is the intensity of the least audible sound. If I0=10^-12 W/m^2 about how many times more intense is a 108 decibel sound than a 100 decibel sound?
A. 4.56
B. 6.31
C. 5.64
D. 7.82

Answer 1

The 108 decibel sound is 6.31 times more intense than a 100 decibel sound option (B) is correct.

What is a logarithm?

It is another way to represent the power of numbers, and we say that 'b' is the logarithm of 'c' with base 'a' if and only if 'a' to the power 'b' equals 'c'.

$\rm a^b = c\\log_ac =b$

We have:

The loudness L of sound in decibels is given by

$\rm L=10log(\dfrac{I}{I_0})$

Here I is the intensity of the sound.

$\rm I_0=10^{-12} \ W/m^2$

Loudness for 108 decibel sound:

$\rm 108=10log(\dfrac{I}{10^-^1^2})$

I = 0.06309

Similarly, for:

100 decibel sound:

i = 0.01

I/i = 0.06309/0.01 = 6.31

I = 6.31i

Thus, the 108 decibel sound is 6.31 times more intense than a 100 decibel sound option (B) is correct.

Learn more about the Logarithm here:

brainly.com/question/163125

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