Note to visitors: Other sites have copied information from this
site. This original method uses a single number (0.301), which was derived
by Dr. Weldon Vlasak, and you will see that it provides a simple way to
estimate the value of a logarithm or a decibel.
I am going to try to show you just how easily logarithms
can be handled and how it is possible to calculate the logarithm of any
number without a table of logs or a calculator using only a single number.
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page for more interesting information in the field of science. I
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For the most part, natural phenomena behave
as exponential functions. The word "exponential" shouldn't scare
anyone because it is simply another way to write a number. The logarithm
of a complex number is much more difficult, but we will deal with real numbers
which are not so difficult to understand as some believe. With the knowledge
gained here, you will then be able to cover any finite range of numbers!
Later we shall also see how decibels are a way to define power levels in
the form of logarithms. Now we will now find the first digit of a logarithm,
which is very easily obtained:
1. Deriving the First Digit of the Logarithm: Since we generally count from one to ten, the number 10 will be the "base"
number (B =10). The log of the number 1 is zero. The log of 10 is the number
"1", which can also be written log(10) = 1. Thus log(100) = 2, log (1000) = 3 and
so on. What could be more simple?
Now let's go in the other direction where we handle
numbers that are less than one. The log of 0.1 is the log (1/10) =
(-1), The log (1/100) = (-2), and the log (1/1000)
= (-3). The base number need not be decimal (base 10), and it
is easy convert any base number, which is something that we will do later.
Adding logarithmic values correlates to multiplication.
The first number of the logarithm of a number N is the number of times the
base number is multiplied. Here are some examples: For base 10 we can write log(10) = 1.0, log(100) = 2.0, and log(1000) = 3.0. We have just
calculated three exact log values of N for these simple cases. In fact,
calculating the log of N was easier than the process of multiplying
or dividing one number by another using the methods of arithmetic. If the
log number is 3, it is simply the number of decimal places for N
= 1000. For numbers less than one, the logarithms are negative and log(0.1) = -1, log(0.01) = -2, and log(0.001) = -3. The intermediate
point is log (1) = 0. Numbers are seldom whole numbers, and there
is a way to handle these numbers, as will be seen later. For example, if
N is between 10 and 100, the log(N) is between 1.0 and 2.0.
2. Now let's do some manipulations using what we have
learned so far. When we add or subtract two log numbers, such as log (10)
+ log(100)] = (1+2) = 3 = log number. This is the same result that
we obtained above where log(1000) = 3, so adding logarithms correlates
to multiplication of real numbers. Again, not so difficult. Similarly
[log(100) - log(10)] = (2 -1) = 1, which correlates to the division of two
numbers, 100 /10= 10, and log(10) = 1. In other words, adding log numbers
equates to the multiplication of two numbers, and subtracting two log numbers
equates to division.
3. Here is an example.Choose a large number that is not a whole
number, say N = 16,777,216. Move the decimal point to where only
one digit is to the left of the decimal point. In this case, the decimal
point is moved seven places to the left. The number of decimal
places that we move to the left is the first digit of the log of N, which
is log (N) = 7.0000. This is the first estimate of the logarithm
of this very large number, and Nest = 10,000,000 = estimated value.
This is within 60% of the value of N, obtained from just
one quick operation! The precise value of the logarithm is log (N) =
7.224719896 = 7 + (log (1.6777216)] and therefore the log error is (
7-7.224719896) = -0.224719896. This is an even lower percentage
error in the log domain - - - of only - 3.11%, but we are not yet done.
When N is less than one, we use a similar
procedure except that the decimal point is moved to the right
until we have just one digit at the left of the decimal point. A
movement of the decimal point to the right is necessary for a number
that is less than one. The log of any finite number can be estimated by
simply moving the decimal point, but this rough estimate of the logarithmic
value may not be sufficiently accurate for various applications. Therefore,
we will now improve the accuracy of the estimate using the method of paragraph 4.
4. A New Binary/Decimal Division Estimation Method:
We will use the the square root of two to our advantage. The following graph, which shows the logarithms of
chosen well-spaced numbers over a range of 10,000:1, is helpful in explaining
the next steps of the procedure. We will use these particular values to
in the graph to derive a fairly accurate logarithm of any number within
that range. The log (N) is plotted below for the following values of N:
(a.) N = 1.414 (the square root of 2),
(b.) N = 2.828 (2 times square root
(c.) N = 5.656 (4 times square root of 2),
(d.) N = 8 (4 times 2)
(e.) N =10, apply to any decade (only
four decades are illustrated). Notice that each the space between numbers
is relatively even, which greatly simplifies the estimation procedure over
Using the above values, it is only
necessary to remember one main number in
order to obtain an accurate estimate of the logarithm of any number! The
points plotted in the above graph include the values of N stated above for
the intermediate points. For example, log (N) = log (2) = 0.301.
.All of the values of the log(N) given above are multiples of either
0.301 or 0.301/2 = 0.1505 (Note that 0.1505 is the
log of the square root of (2). So for this procedure,
you only need to remember this number:
For the specific values N of 1, 2, 4, and 8
in the above graph, the logarithmic value increases consecutively by 0.301 within each
decade (0, 0.301, 0.602, and 0.903) = 0.301
x (0, 1, 2, 3). For instance, log (8) = 3 x log(2) = 3 x 0.301 =
0.903. We now have four binary-encoded decimal points in each
decade, and a total of 16 additional points for the entire graph. Also,
observe that the log of the square root of two is (0.301/2 = 0.1505),
which allows us to obtain intermediate values between any two adjacent
binary-value points on the above curve. This is not an exact interpolation
method, but it is pretty good, and we can easily make it more accurate.
We can also divide numbers in a similar way in order to determine the log value. Choosing N = 10, if we divide N by two we get the number
five. Again, log (N) = 1, but in this case, we divide N by two and we
subtract the log value resulting in log (N/2) = log (5) = (1 - 0.301)
= 0.699. The actual value is 0.69897... Similarly, log (N/4) = log
(2.5) = (0.699 - 0.301) = 0.398. This gives us more points on
the above graph if we choose to use them for greater accuracy or easier
So now we know the log values of 10, 5 and 2, which are 1.0,
0.699 and 0.301. The log of 4 is therefore 0.602, and
the log of 8 is 0.903. For numbers less than one, the log of 0.5 is -0.301,
the log of 0.25 is -0.604, and the log of 0.125 is -0.903.
All of this from just remembering one number (0.301), and by multiplying
these numbers by multiples of ten, we just add integers. Similarly, by dividing
numbers by ten we subtract multiples of ten. Thus we can determine the log
values of an extremely wide range of numbers to a reasonable degree of accuracy
without a table of logarithms, a computer or a calculator!
With the values of the points on the above curve
we can also utilize and interpolation method. For instance, the let us consider
N = log (2 x sqrt 2) = log (2.828). Using the above method, log
(2.828) = log (2) + log (sqrt 2) = (0.301 + 0.1505)) = 0.4515. This
is similar to the method of the previous paragraph where we determined that
log 8 = log (2 x 2 x 2) = (0.301 x 3) = 0.903, In order to obtain
the closest estimate for the log of any number, choose the closest convenient
values to the number (N) for which you want to find the logarithm
as pictured in the above graph.
5. Using This Interpolation Method for Greater Accuracy
in the Estimate of paragraph 2: Recall that N
= 16,777,216, and that the estimate number was 10,000,000 for a log value of 7. We now add 0.301 for a new log value of 7.301 and the new estimate is 14,142,136. Our new estimate is now within 16% of the log value.If we now observe the points on the graph, 1.677 is about half
way between 1.414 and 2, so we now divide 0.1505 (the
log of the square root of two) in half and add it to the previous estimate,
resulting in 7.1505 +0.07525 = 7.22575, resulting in an error of
only Error = - 0.2968%. In most cases where logarithms are used in
engineering and science, this type of linear interpolation will be more than sufficient.
6. The Logarithm of a Binary Number: Now we
will choose a different base number B = 2 (binary number system),
which in this case the number of times that the number 2 is multiplied.
For N = 16, the number two is multiplied four times, N = 2 x 2
x 2 x 2 = 16 = 2^4, and its base 2 logarithm is:log (16) =
4. Similarly, for N = square root of 2 = 1.4142,
the (binary) log value is 0.5, whereas the (decimal) log value is
0.301 for N =2.
Decibels: A decibel value refers to a ratio of two numbers. The decibel
(abbreviated dB) was originally defined in terms of audio power,
and an increase in power level of one decibel was the least amount of audio
power that the average human ear could hear. We now use decibels to define
signal levels over a very wide range for electrical signals, etc. The most
common reference point for zero dB is one milliwatt of power. Once the reference
value is known, the absolute power can be expressed by a logarithmic value.
The decibel is defined as 10 log (P2/P1). Therefore, an increase
in power by 10:1 is 10 log(10) = 10 x 1 = 10 dB. Going in
the other direction, an increase of 10 dB is 10 log (P2/P1)
= 10, so log (P2/P1) = 1.0, which is 10^1 = 10. Now
consider an increase in power of 3 dB, in which case 10 log (P2/P1)
= 3, and thus log (P2/P1) = 0.300, which is sufficiently close
to our magical number of 0.301 for the logarithmic value that corresponds
to an increase of 3 dB. Therefore, 3 dB means an increase in power
of 2:1 (it will help to remember this). The term dBm means
that the reference level is one milliwatt, so in this case 3 dBm
means a power level of 2 milliwatts.
8. A Simple Exercise: In order to test
what you have learned here, consider the following problem:
a.)Calculate the relative increase in power for an
increase of one dB.
b.) Determine the absolute power level for one dBm.
See the answers
to the above exercises.
See a plot
of decibels and Q&A.
Continue: See more plots
If you find any difficulties in using the methods
outlined here, or if you have suggestions for improving the clarity of the
presentation, please email
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