Question

The decibel rating D is related to the sound intensity I by the formula D=10log(I/(10^-16)) for the noise in decibels. a) let D and d represent the decibel ratings of sounds of intensity I and i, respectively. Using properties of logarithms, find a simplified formula for the difference between the two ratings, D-d, in terms of the two intensities I and i. b) If a sounds intensity quadruples, how many decibels louder does the sound become?

Answer 1

The decibel D is related by the formula:

$D=10\log _{10}(\frac{l}{10^{-16}})$

Now, let's find a simplified formula between the two rating D and I using properties of logarithms:

We are going to use the same formula but changing D by d and I by i, then subtract both formulas:

$d=10\log _{10}(\frac{i}{10^{-16}})$

D-d:

$D-d=10\log _{10}(\frac{I}{10^{-16}})\text{ - 10}\log _{10}(\frac{i}{10^{-16}})$

Factorize the number 10:

$D-d=10\lbrack\log _{10}(\frac{I}{10^{-16}})-\log _{10}(\frac{i}{10^{-16}})\rbrack$

Use the next rule of logarithms:

$\log _a(\frac{m}{n})=\log _am-\log _an$

So:

$D-d=10\lbrack\log _{10}I-\log _{10}10^{-16}-\log _{10}i+\log _{10}10^{-16}\rbrack$

Operate the common terms:

$D-d=10\lbrack\log _{10}I-\log _{10}i\rbrack$

Now, we are going to use the same rule presented before by changing the rest by a division:

$D-d=10\log _{10}(\frac{I}{i})$

With the before formula you can solve the difference between two ratings.

b)The sound intensity now quadruples, using the first given formula find the decibels louder:

So I = 4, replace this value and solve:

$D=10\log _{10}(\frac{4}{10^{-16}})$

We use the same property changing the division by a subtraction:

$D=10\log _{10}4-10\log _{10}10^{-16}$

Now, we are going to use the next property:

$\log _{}a^b=b\cdot\log _{}a$$D=10\log _{10}4-(10\cdot-16\log _{10}10)$