*Understanding Logarithms and Decibels*

*Solve a logarithm without a calculator or table*

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.

Before you leave the web site, please check out ourI now reside inmain pagefor more interesting information in the field of science. I try to make science simple to understand, as you will see., so use the contact page for help with local discussion groups or schools or for questions..NebraskaFor the most part,

natural phenomenabehave 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 ismoved 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 islog (N) = 7.0000. This is the first estimate of the logarithm of this very large number, andNest = 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 islog (N) = 7.224719896 = 7 + (log (1.6777216)]and therefore the log error is ( 7-7.224719896) =-0.224719896. This is an even lower percentage errorin 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 ismoved to the rightuntil we have justone digitat the left of the decimal point. A movement of the decimal pointto the rightis 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 estimateusing 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 of 2),(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 any range!

Using the above values, it is only necessary to remember one main numberin 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 either0.301or0.301/2 =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:

0.301For the specific values N of 1, 2, 4, and 8 in the above graph, the logarithmic value increases consecutively by

0.301within 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

dividenumbers in a similar way in order to determine the log value. ChoosingN = 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 wesubtractthe log value resulting inlog (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 calculation.So now we know the log values of

10, 5 and 2, which are1.0, 0.699 and 0.301. Thelog of 4is therefore0.602, and thelog of 8 is 0.903. For numbers less than one, thelog of 0.5 is -0.301, thelog of 0.25 is -0.604, and thelog 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 thatlog 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 thatN = 16,777,216, and that the estimate number was10,000,000for a log value of7. We now add 0.301 for a new log value of7.301and the new estimate is14,142,136. Our new estimate is now within 16% of the log value.If we now observe the points on the graph,1.677is about half way between1.414and2, so we now divide0.1505(the log of the square root of two) in half and add it to the previous estimate, resulting in7.1505 +0.07525 = 7.22575, resulting in an error of onlyError = - 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 numberB = 2(binary number system), which in this case the number of times that the number2is multiplied. ForN = 16, the number two is multiplied four times,N = 2 x 2 x 2 x 2 = 16 = 2^4, and itsbase 2logarithm is:log (16) = 4. Similarly, forN =square root of 2= 1.4142, the (binary) log value is0.5, whereas the (decimal) log value is0.301forN =2.

7.Understanding Decibels: A decibel value refers to a ratio of two numbers. The decibel (abbreviateddB) 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 as10 log (P2/P1). Therefore, an increase in power by10:1is10 log(10) = 10 x 1 = 10 dB. Going in the other direction, an increase of10 dBis10 log (P2/P1)= 10, solog (P2/P1) = 1.0, which is10^1 = 10. Now consider an increase in power of3 dB, in which case10 log (P2/P1) = 3, and thuslog (P2/P1) = 0.300, which is sufficiently close to ourmagical number of 0.301for the logarithmic value that corresponds to anincrease of 3 dB. Therefore, 3 dB means anincrease in power of(it will help to remember this). The term2:1dBmmeans that the reference level isone milliwatt, so in this case3 dBmmeans a power level of2 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

answersto the above exercises.

See aplot of decibels and Q&A.

Continue:See moreplots and answers.If you find any difficulties in using the methods outlined here, or if you have suggestions for improving the clarity of the presentation, please

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