LOGARITHMS |
The logarithm is perhaps the single, most useful arithmetic concept in all the sciences; and an understanding of them is essential to an understanding of many scientific ideas. Logarithms may be defined and introduced in several different ways. But for our purposes, let's adopt a simple approach. This approach originally arose out of a desire to simplify multiplication and division to the level of addition and subtraction. Of course, in this era of the cheap hand calculator, this is not necessary anymore but it still serves as a useful way to introduce logarithms. The question is, therefore:
Is there any operation in mathematics which produces a multiplication by the performance of an addition?
With not too much thought, the answer should come to you.
What is 2^{3} x 2^{4}.
The answer is 2 ^{7} which is obtained by adding the powers 3 and 4. This is correct, of course, since 2^{3} x 2^{4} is just seven 2s multiplied together. Note that this addition trick does not work for the case of 3^{3} x 2^{4}. The base numbers must be the same, as in the first case, where we used 2.
In general, this addition trick can be written as p^{a} x p^{b} = p^{a+b}. This expression will do our job of multiplying any two numbers, say 1.3 and 6.9, if we can only express 1.3 as p^{a} and 6.9 as p^{b}.
What number will we use for the base p? Any number will do, but traditionally, only two are in common use:
Ten (10) and the transcendental number e (= 2.71828...), giving logarithms to the base 10 or common logarithms (log), and logarithms to the base e or natural logarithms (ln).
If you would like to know why this strange number e is used click here.
Let's first talk about logarithms to the base 10 or common logs. We thus choose to let our number 1.3 be equal to 10^{a}.
`a' is called "the logarithm of 1.3". How large is `a'? Well, it's not 0 since 10^{0} = 1 and it's less than 1 since 10^{1} = 10. Therefore, we see that all numbers between 1 and 10 have logarithms between 0 and 1. If you look at the table below you'll see a summary of this.
Number range 1 - 10 or 10^{0} - 10^{1} 10 - 100 or 10^{1} - 10^{2} 100 -1000 or 10^{2} - 10^{3} etc. |
Logarithm Range 0 -1^{ } 1 - 2^{ } 2 - 3^{ } etc. |
I'm going to take a moment to discuss your calculator. If you don't have a calculator with scientific functions on it, you should get one before proceeding in this tutorial since the rest depends on it.
Most calculators are very straightforward in obtaining the logarithm. They either have two logarithm keys or a dual function key. In any case, the labels will be `log' and `ln' which is often pronounced `lon'.
Log is the key for logs to the base 10 and ln is for natural logs. We want logs to the base 10 in our example so we use `log'. Enter 1.3 on your calculator, and then press the log key.
Do you have 0.113943? You should have. This number then is `a' back in our previous expression and therefore the logarithm of 1.3. Pause now and determine `b' in that expression, the logarithm of 6.9.
You should have 0.838849 for the log of 6.9. If not, review what we have done and try again. Now we are going to do something silly in view of the fact that you have a calculator. We're going to use the two logarithms you have evaluated to find the product of 1.3 and 6.9. Of course, you can do it quickly with your calculator, but this will show that logarithms do what they are supposed to do. According to our original idea, the sum of the two logarithms was supposed to be the logarithm of the answer.
Now add the two logarithms. The sum is 0.952792. This is the logarithm of the answer. If we only knew what number had 0.952792 as its logarithm, we would know the value of 1.3 x 6.9.
The problem of finding a number when you know its logarithm is called finding the "antilogarithm" or sometimes "exponentiation". Again, lets look at your calculator. Here is where calculators differ a lot and I hope I mention one that is something like yours.
You should look for a key on your calculator that says something like 10^{x} or 10^{y}. If so, then pressing that key will take the antilog of the number in the display. Alternatively, your calculator may have an "inverse" key. If so, then pressing inverse and then log will take the antilog of the number in the display. Enter 0.952792 into your calculator and find the antilog.
Did you get 8.97? If not, try again. Of course, you might have got something like 8.96999 but of course that really is 8.97. Now, multiply 1.3 x 6.9 on your calculator and you'll see that 8.97 is indeed the correct answer.
The whole operation could be done with natural logarithms as well as shown below.
1.3 x 6.9 = ?
ln 1.3 = 0.262364^{ } ln 6.9 = 1.931521^{ } total = 2.193885^{ } |
i.e. 1.3 = e^{0.262364} i.e. 6.9 = e^{1.931521} i.e. 1.3 x 6.9 = e^{2.193885} |
antiln 2.193885 = 8.97
If the sum of logarithms gives the product of two numbers, then the difference gives the quotient.
In the table below, I've taken the difference between the ln of 1.3 and the ln of 6.9. Check it on your calculator.
1.3/6.9 = ?
ln 1.3 = 0.262364 ln 6.9 = 1.931521 ln 1.3/6.9 = -1.669157 |
If you didn't get 0.1884, try again. Of course, this is just 1.3 divided by 6.9. In the table below, I have done the whole problem over again using common logs. Pause here and check it.
log 1.3 = 0.113943 log 6.9 = 0.838849 log(1.3/6.9) = -0.724906 antilog (-0.724906) = 0.1884 |
The logarithmic and exponential functions are very important since many physical and biological processes can be described by them.
For example, suppose you have a certain number of radioactive atoms at time t = 0. Let's let this number be N_{0}. Radioactivity behaves in such a way that the number N of radioactive atoms remaining at a later time t is given by a linear variation of the logarithm of N with t.
That is, a graph of ln N vs t is a straight line. You know that the equation of such a straight line is given by y = mx+b where m is the slope and b is the y intercept. Therefore, the equation of radioactivity is ln N = -kt + ln N_{0} where ln N_{0} is the y intercept and the slope of the line is -k. |
Let's now examine the equation we derived
for radioactivity, ln N = ln N_{0} -kt. Here is
where it is important to be able to do algebra with logarithms.
Lets get the logarithms on one
side so that we get
Quiz Try the following questions:
1. ln (7.42) = ?
2. log (7.42) = ?
3. ln (e^{kt}) = ?
4. log (e^{kt}) = ?
5. antilog 0.8704 = ?
6. antiln 2.0042 = ?
7. 10^{0.8704} = ?
8. e^{2.0042} = ?
Return to Tutorial Main Page