Back
to "Answer-This" Listing

**"Answer This" Questions**

**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 10^{6}
h) 0.00032
i) 438x10^{5}
j) 5x10^{–3}/2x10^{3}

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(10^{3})
= 3.

b) From the log table: log(8) = log(10^{0.903}) = 0.903

c) From the log table: log(4.8) = log(10^{0.681}) = 0.681

d) Use the table to find the log of 3.5: log(3.5 x 10^{–2})
= log(10^{0.544}x 10^{–2}) = log(10^{0.544–2})
= –1.456

e) log(2000) = log(2 x 10^{3}) = log(10^{0.301}x 10^{3})
= log(10^{0.301+3}) = 3.301

f) log(500,000) = log(5 x 10^{5}) = log(10^{0.699}x
10^{3}) = log(10^{0.699+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 10^{6}) = log(8.4
x 10^{4}) = log(10^{0.924 }x 10^{4}) = log(10^{0.924+4})
= 4.924

h) Convert to scientific notation first: log(0.00032) = log(3.2 x 10^{–4})
= log(10^{0.505–4}) = –3.495.

i) log( 438x10^{5}) = log(4.38x10^{7}) = log(10^{0.641}x
10^{7}) = 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}/2x10^{3}) = log(2.5
x 10^{–6}) = log(10^{0.398} x 10^{–6})
= log(10^{0.398–6}) = –5.602

**Part
2:**

a) If x is the logarithm, then the number (antilog) is 10^{x}.

b) 10^{2} = 100.

c) 10^{4.8} = 10^{0.8} x 10^{4} = 6.3 x 10^{4}.
In the last step we found the number in the table whose logarithm is closest to 0.8.

d) 10^{0.5} = 3.16. This is the square root of 10.

e) 10^{–0.5} = 1/10^{0.5} = 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} = 10^{1–1–0.5}
= 10^{0.5} x 10^{–1} = 3.16 x 10^{–1}
= 0.316

f) Use the second method of e) here: 10^{–16.5} = 10^{17–17–16.5}
= 10^{0.5} x 10^{–17} = 3.16 x 10^{–17}
.

____________________________________

**11-H.
****Decibels.**

How
many decibels is a sound having an intensity of 5 x 10^{–8}W/m^{2}?

What is the intensity in W/m^{2} of an 89 dB sound?

**Answer:**

**Part
1**: I/I_{min} = 5 x 10^{–8}/10^{–12}=
5 x 10^{4} = 10^{4.7}. The exponent 4.7 is the log of
the previous number. The decibel level is then 47 dB.

**Part
2**: I/I_{min} = 10^{8.9} = 10^{8+0.9} = 7.9
x 10^{8}. Thus *I* = *I*_{min} x 7.9 x 10^{8}
= 10^{–12} W/m^{2} x 7.9 x 10^{8} =7.9
x 10^{–4} W/m^{2}.

____________________________________

**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.

____________________________________

**11-J.
****Combined
sound sources.**

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

**Answer:**

5 I/I_{min}
= 5x10^{-2}/10^{-12} = 5x10^{10} = 10^{10.7}.
The decibel level is then 107 dB.