# Logarithm word problem, is there enough info?

1. Feb 23, 2017

### late347

1. The problem statement, all variables and given/known data
Intensity of sound is (W/m^2) and is inversely proportional to the the square of the distance measured from the sound source. The noise level of a risiing jet aircraft at the distance of 30m, is 140 dB.

How far from the jet aircraft is the noise level at the level of 120 dB (which is about the noise of rock concert)

2. Relevant equations
there were no equations provided, and this was from a math textbook. I think the this is the entire problem statement for the specific textbook problem.

I was wondereing how it could be solved with only the information in the problem statement.

However in an earlier problem in the same textbook there was a formula for noise level (L, which has decibels) I'm not 100% certain that is the right word to use in English for the quantity, which is measured in decibels.

L = 120 + 10 lg(I) where I is the sound's intensity.

certainly it seems the problem was solvable if you used that formula. You have to put it in the form of I= something.

3. The attempt at a solution

we know essentially that there is inverse relationship therefore k = constant which we can calculate initially using that L= whatever fromula
$k= I*r^2 \nonumber \\ \leftrightarrow ~~k=10^{\frac{140-120}{10}}*30^2 \nonumber \\ \leftrightarrow ~~k=10^2*30^2 =90000$

from earlier problem's formula we can get an idea what L = noise level (???) will be and what Intensity will be
$L= 120 +10lg(I) \nonumber \\ \leftrightarrow ~~ I= 10^{\frac{L-120}{10}} \nonumber$

original formula can be used and intensity can be plugged into that original formula

$r^2= \frac{k}{I} \nonumber \\ \leftrightarrow ~~r^2=\frac{90000}{10^{\frac{L-120}{10}}} \nonumber \\ \leftrightarrow ~~r^2=\frac{90000}{10^{\frac{120dB-120dB}{10}}} \nonumber \\ \leftrightarrow ~~r^2=\frac{90000}{1} \nonumber \\ \leftrightarrow ~~r=300$

I was wondering if there is alternative way to do it.

2. Feb 23, 2017

### BvU

There is. 140 dB at 30 m and it falls off with the square of the distance. 20 dB is a factor ... ? So the distance has to increase by a factor √...

3. Feb 23, 2017

### late347

I got lost there I'm afraid. 20*7 would be factors of 140 I guess...

4. Feb 23, 2017

### BvU

1 Bel is a factor of 10 so 20 dB is a factor of 100.

5. Feb 23, 2017

### late347

I have no idea how decibel scale works I'm sorry... This homework problem was not a physics problem originally because it was in a math textbook first of all... But the moderators of course moved the problem into introductory physics homework subforum.

It seems I lack a deep understanding of the how the decibel scale works when you increase the distance or decrease the distance. I know how to use a particular formula though, such as this one

$L= 120 + 10 lg( I )$

I know from my physics class that sound's intensity is denoted by capital I, and its unit is the Watt/m^2
and intensity follows the inverse square law.

$I= \frac{Power}{4πr^2}$

However that ties into decibels I don't really know for sure...

6. Feb 23, 2017

### BvU

7. Feb 25, 2017

### late347

Well, I did some googling and according to hyperphysics http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/db.html#c3
http://hyperphysics.phy-astr.gsu.edu/hbase/Acoustic/isprob.html#c3

there should be a way to get a comparison equation such as $\frac{I_1}{I_2} = \frac{(r_2)^2}{(r_1)^2}$

Let $I_1 =$ the intensity at 30m and at 140dB
Let $I_2 =$ the intensity at unknown meters and at 120dB
Of course $r_1= 30m$
from decibel definition we can calculate from two separate calculations what the intensities $I_1$ and $I_2$ will be, but we don't know these yet I think (unless you're a physics wizard at this sound business...)

I still did not quite get what you meant by the "something is a factor of something" thing, but with the hyperphysics decibel definition and the intensities ratio = ´( inverted distances ratio)^2 you could attempt ¨the same prob from the decibel definitoin equation and plug in the now solved for intensities into the ratio= inverted ratio squared. We already had proven at physics class the intensities ratio thing. I suppose the comparison of ratios assumes that the sending power is the same

Decibel definition was given such that
$L (dB) = 10*lg(\frac{I}{I_0})$

fruthermore we know that I_0 will be the same value at $I_0 = 10^{-12} \frac{W}{m^2}$

start solving Intensity´_1

$140 = 10 lg (I_1/10^{-12})$
$\leftrightarrow 14 = lg (I_1/10^{-12})$
$\leftrightarrow 14 = lg(I_1) - lg(10^{-12})$
$\leftrightarrow 14 = lg (I_1) +12$
$\leftrightarrow 2 = lg (I_1)$
$\leftrightarrow 10^2 = I_1$

start solving Intensity´_2

$120 = 10 lg (I_2/10^{-12})$
$\leftrightarrow 12 = lg (I_2/10^{-12})$
$\leftrightarrow 12 = lg (I_2) - lg(10^{-12})$
$\leftrightarrow 12 = lg (I_2) - (-12)$
$\leftrightarrow 0 = lg (I_2)$
$\leftrightarrow 10^0= I_2$

then we get the comparing ratio of intensities to ratio of square of distances

$\frac{100}{1} = \frac{{r_2}^2}{30^2}$
$\leftrightarrow 100 * 900= {r_2}^2$
$\leftrightarrow 90 000= {r_2}^2 ,~~ take ~~root$
$\leftrightarrow 300m= {r_2}$

8. Feb 25, 2017

### BvU

Understandable: I'm dealing with this logarithmic scale in a rather casual way -- assuming logarithms are familiar to everyone.

The whole basis is that $\log (a*b) = \log a + \log b$ and thereby $\log (a/b) = \log a - \log b$. So a difference on a log scale is a ratio (a factor) on the scale of the numbers $a$ and $b$.

And yes, you should assume that the sending power is the same; there's no exercise without that assumption

One good tip to keep in mind: in math and physics (and in fact all over science ), you can only exponentiate or take logarithms of dimensionless numbers. In general that's a nice check to see if your algebra isn't off the rails.

9. Feb 25, 2017

### late347

If you had some other reasoning or solution I think youre welcome to post it.

I think I tried my best with the problem especially the second attempt with using physics principles to get the result (though this entire thing was indeed a math textbook problem)

I already asked my stepfather about this problem (who studied math in university) but I guess he couldnt figure out how to do this problem purely using "math principles" (he never studied much physics)

10. Feb 25, 2017

### BvU

I'm lost as to where you are experiencing difficulty: is it in the math (which I tried to address with this latest posting) or did I miss something and does it sit somewhere else ?

11. Feb 25, 2017

### late347

By what reasoning do you arrive to the conclusion that the correct distance should be 300m when the decibel level falls from 140dB(at 30m) to 120dB at 300m as it turned out.

Well, of course now that I got the correct result in some semi-logical manner, this means that you could get 300m by multiplying $\sqrt(100)*30m=300m$

I tried reading the wikipedia English article about decibels, but I didn't understand anything really because the English terms seem to be different than the Finnish ones. This happened at e.g. the definitions part https://en.wikipedia.org/wiki/Decibel#Definition
I tried switching to the Finnish language wiki article but it was only a stump article. hyperphysics seemed to be a good source though.

I just could not piece together your bolded advice at all. This may or may not be worrying as I have a math exam on monday. I reckon that this type of problem would have been on the difficult scale of problems, considering the types of logarithm problems we had done at school during our course.

I started out from the physics principle for example, and then I computed the correct intensities of sound, by which I calculated the corresponding ranges, and got 300m that way. I even put in good written descriptions and defined the variables, and I used the equivalency signs $\leftrightarrow$

12. Feb 25, 2017

### BvU

As far as I can see you did just fine. Still don't see you problem, sorry. In the link you see $L_p = 10 \log_{10} \left (P\over P_0\right ) dB$ Your reference power is 140 $dB$ at 30 m. With $L_p=120 -140 = -20$ you get ${P\over P_0} = 10^{-2}$ and with ${P\over P_0} = {r_0^2\over r^2}$ the equation becomes ${r^2} = {r_0^2\over 10^{-2}}$. That's all.

Don't let it worry you.
Advice for tomorrow: switch your mind to the position of the question composer from time to time: what can they ask of you ?

13. Feb 26, 2017

### late347

I thought reference power meant the wattage not decibel (clue: Power = P = Watt).

I mean, it would not make much sense if it was wattage, but I think our physics and math teachers taugth that decibel is a unit of noise level ($dB$). And we were only taught about noise level in decibels and the intensity of sound was another quantity, and the ratio of the intensities can be put inside the log_10() function ( $\frac{Watt}{m^2}$)