Question

The decibel level of a sound is given by the equation D = 10log10 (I/I_0) where I is the sound's intensity and I0 corresponds to the intensity of the weakest sound that can be detected by the human ear. A sound of 140 decibels produces pain in the average human ear. Approximately how many times greater than I0 must the intensity I of a sound be to reach this decibel level?

Answer 1

a. 22026 times greater
b. 14 times greater
c. 1.14612 times greater
d. 2.63906 times greater
e. 1014 times greater

Answer 2

@zepdrix @Compassionate Help plz?!

Answer 3

Opps, I think it ment I_0, not 10

Answer 4

So our function, decibels as a function of sound intensity,\[\Large\bf\sf \mathcal D\quad=\quad 10\log\left(\frac{\mathcal I}{\mathcal I_o}\right)\]

Answer 5

And they want us to compare I to I_o when D=140

Answer 6

\[\Large\bf\sf 140\quad=\quad 10\log\left(\frac{\mathcal I}{\mathcal I_o}\right)\]

Answer 7

So we want to start by solving for \(\Large\bf\sf \dfrac{\mathcal I}{\mathcal I_o}\)

How can we do that? :)

Answer 8

Do we multiply both sides

Answer 9

Umm let's divide each side by 10, that will clean things up a little bit.

Answer 10

\[\Large\bf\sf 14\quad=\quad \log\left(\frac{\mathcal I}{\mathcal I_o}\right)\]

Answer 11

From here you can either `exponentiate` or simply rewrite it in exponential form if you remember how.

Answer 12

Oh sry, I meant to put 10^14 not 1014 on the choice e.

Answer 13

I don't know how to do tht

Answer 14

So we'll try exponentiating.
It's a bit of a weird process but worth getting comfortable with.

Rewrite each side as an exponent of base 10,\[\huge\bf\sf 10^{14}\quad=\quad 10^{\log\left(\frac{\mathcal I}{\mathcal I_o}\right)}\]

Answer 15

Since log base 10, and the exponential function base 10 are `inverse functions` of one another.

They "undo" each other.
You can think of it as "cancelling out" if that's easier to remember.
It's not as accurate, but that's ok.\[\Large\bf\sf 10^{14}\quad=\quad \left(\frac{\mathcal I}{\mathcal I_o}\right)\]

Answer 16

Then multiply I_o to the other side,

Answer 17

\[\Large\bf\sf 10^{14}\mathcal I_o\quad=\quad \mathcal I\]