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Chapter 11

11-G. Logarithms

Find the logarithms of the following numbers.
a) 1,000    b) 8    c) 4.8    d) 3.5 x 10–2     e) 2000     f) 500,000
g) 2.1 x 10–2 x 4.0 x 106
   h) 0.00032    i) 438x105    j) 5x10–3/2x103

The following are logarithms (the power of ten representation) of the some numbers. Find the numbers.
a) x
   b) 2    c) 4.8    d) 0.5    e)–0.5    f)–16.5

 

Answer:

Part 1:
a) Convert to scientific notation first: log(1,000) = log(
103) = 3.
b) From the log table: log(8) = log(100.903) = 0.903
c) From the log table: log(4.8) = log(100.681) = 0.681
d) Use the table to find the log of 3.5: log(3.5 x 10–2) = log(100.544x 10–2) = log(100.544–2) = –1.456
e) log(2000) = log(2 x 103) = log(100.301x 103) = log(100.301+3) = 3.301
f) log(500,000) = log(5 x 105) = log(100.699x 103) = log(100.699+5) = 5.699
g) The log of a product of terms is the sum of the logs of the terms, but it is simpler to do the multiplication before taking the log: log(2.1 x 10–2 x 4.0 x 106) = log(8.4 x 104) = log(100.924 x 104) = log(100.924+4) = 4.924
h) Convert to scientific notation first: log(0.00032) = log(3.2 x 10–4) = log(100.505–4) = –3.495.
i) log( 438x105) = log(4.38x107) =  log(100.641x 107) = 7.641
j) The log of a ratio of terms is the log of the numerator minus the log of the denominator, but it is simpler to do the division before taking the log: log(5x10–3/2x103) = log(2.5 x 10–6) = log(100.398 x 10–6) = log(100.398–6) = –5.602

Part 2:
a) If x is the logarithm, then the number (antilog) is
10x.
b) 102 = 100.
c) 104.8 = 100.8 x 104 = 6.3 x 104. In the last step we found the number in the table whose logarithm is closest to 0.8.
d) 100.5 = 3.16. This is the square root of 10.
e) 10–0.5 = 1/100.5 = 1/3.16 = 0.316. Another way to proceed is to add (and also subtract) a large enough positive integer in the exponent to change the negative decimal exponent to a small positive decimal: 10–0.5 = 101–1–0.5 = 100.5 x 10–1 = 3.16 x 10–1 = 0.316
f) Use the second method of e) here: 10–16.5 = 1017–17–16.5 = 100.5 x 10–17 = 3.16 x 10–17 .

 

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11-H. Decibels.

How many decibels is a sound having an intensity of 5 x 10–8W/m2?
What is the intensity in W/m2 of an 89 dB sound?

Answer:

Part 1: I/Imin = 5 x 10–8/10–12= 5 x 104 = 104.7. The exponent 4.7 is the log of the previous number. The decibel level is then 47 dB.

Part 2: I/Imin = 108.9 = 108+0.9 = 7.9 x 108. Thus I = Imin x 7.9 x 108 = 10–12 W/m2 x 7.9 x 108 =7.9 x 10–4 W/m2.

 

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11-I. Phons.

What is the intensity level in dB of a 200 Hz tone with a loudness level of 50 phons?
What is perceived as louder, a 100 Hz tone played at 51 dB or a 10,000 Hz tone played at 30 dB?

 

Answer:

Part 1: To answer the question, look at the 50 phon contour on the Fletcher-Munson diagram and find where it meets the vertical line corresponding to 200 Hz; you now read the decibel level off the left axis to be very near 60 dB.

Part 2: On the Fletcher-Munson diagram, find the 51 dB intensity level (read it off the left-most vertical axis) and move horizontally across to the 100 Hz vertical line. You can see that this point is on the 20 phon contour. The 30 dB intensity level from the left axis also intersects the 20 phon contour when you move over to 10,000 Hz. Thus these two sounds are equally loud.

 

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11-J. Combined sound sources.

Five sound sources, each at an intensity of 10–2 W/m2, are combined. What is the dB level of the combined sounds?

 

Answer:

5 I/Imin = 5x10-2/10-12 = 5x1010 = 1010.7. The decibel level is then 107 dB.

 

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