Logarithms Tutorial
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 \(2^3 \times 2^4\).
The answer is \(2^7\) which is obtained by adding the powers 3 and 4. This is correct, of course, since \(2^3 \times 2^4\) is just seven 2s multiplied together. Note that this addition trick does not work for the case of \(3^3 \times 2^4\). 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 \(p^a \times p^b = p^{a+b}\). 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 (\(ln\)).
The two bases for logarithms in common use are a) \(10\) and b) the transcendental number \(e = 2.71828---\)
- a) For ordinary computations, logarithms to the base \(10\) are most common. This is the common or Briggsian system first devised by Henry Briggs (1560-1631) with the assistance of Napier.
- b) Logarithms to the base e are called Naperian or natural logarithms for their inventor John Napier (1550-1617). For general numerical computation they have no advantage over common logarithms. Their origin and usefulness must be sought in more sophisticated mathematics than just multiplication and division of numbers and is connected to the special properties of the number e. Some of these properties are:
1. The exponential function ex is the only one that has itself as a derivative i.e.,
\(de^x/dx = e^x\)
2. The simple but very important differential equation expressing constant percentage growth with time \(t\)
\(dx/x = (const)dt\)
has the solution
\(x = x_0e^{(const)t}\)
or taking natural logarithms
\(ln(x/x_0) = (const)t\)
3. The numerical values of logarithms are determined by summing the terms of an infinite series. The series for \(e\) is
\(e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ----\)
[Remember \(n!\) is n-factorial or \((n)X(n-1)X(n-2)X---(1)\)]
For e raised to the power \(x\) it is
\(e^x = 1 + x + x^2/2! + x^3/3! + ---\)
Evaluating numbers in this way is labour intensive so series that converge rapidly are the best. The most rapidly converging series is that for \(ln\) (i.e., to the base \(e\)). Any other base logarithm (e.g., to base \(b\)) can be found from the simple relation
\(\log _ba = ln\; a/ln \; b\)
Therefore once the most efficient set of logarithms have been found any others follow from a simple ratio.
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 \(10a\).
\(1.3 = 10a\)
`\(a\)' is called "the logarithm of \(1.3\)". How large is `\(a\)'? Well, it's not \(0\) since \(100 = 1\) 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 | Logarithm Range |
|---|---|
| \(1 - 10\) or \(10^0 - 10^1\) | \(0 - 1\) |
| \(10 - 100\) or \(10^1 - 10^2\) | \(1 - 2\) |
| \(100 -1000\) or \(10^2 - 10^3\) | \(2 - 3\) |
| etc. | etc. |
You see, we have the number range listed on the left and the logarithm range listed on the right. For numbers between \(1\) and \(10\), that is between \(10^0\) and \(10^1\), the logarithm lies in the range \(0\) to \(1.\) For numbers between \(10\) and \(100\), that is between \(10^1\) and \(10^2\), the logarithm lies in the range \(1\) to \(2\), and so on. Now in the bad old days before calculators, you would have to learn to use a set of logarithm tables to find the logarithm of our number, \(1.3\), that we asked for earlier. But nowadays, you can get it at the press of a button on your calculator.
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 \(1.3 \times 6.9.\)
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 \(1.3 \times 6.9\) 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.
\(1.3 \times 6.9 = ?\)
\(ln 1.3 = 0.262364\) i.e. \(1.3 = e^{0.262364}\)
\(ln 6.9 = 1.931521\) i.e. \(6.9 = e^{1.931521}\)
\(total = 2.193885\) i.e. \(1.3 \times 6.9 = e^{2.193885}\)
\(antiln 2.193885 = 8.97\)
If the sum of logarithms gives the product of two numbers, then the difference gives the quotient.
In the instance below, I've taken the difference between the ln of \(1.3\) and the ln of \(6.9.\) Check it on your calculator.
\(1.3/6.9 = ?\)
\(ln 1.3 = 0.262364\)
\(ln 6.9 = 1.931521\)
\(ln 1.3/6.9 = -1.669157\)
\(antiln (-1.669157) = 0.1884\)
Don't be afraid of the negative sign. It simply means that the answer is less than \(1.\) Enter \(-1.669157\) on your calculator, then find its antiln. Note we are working with natural logs in this example.
If you didn't get \(0.1884\), try again. Of course, this is just \(1.3\) divided by \(6.9.\) In the instance 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 N0. 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
\(ln N - ln N_0 = -kt.\)
But we know that the difference of logarithms is the logarithm of the quotient so the left-hand side becomes ln \(N/N_0\). Now let's take antilogarithms and do the right side first. The antiln of any quantity is the number \(e\) to the power of that quantity so the right-hand side becomes \(e^{-kt}\). The left-hand side is the antiln of the ln and so it just becomes \(N\) divided by \(N_0\). Finally, we can rearrange to put the final equation in the form
\(N = N_0 e^{-kt}\)
which is called the equation of "exponential decay", so you can see why taking an antiln is often called "exponentiation".
Quiz
- \(ln (7.42) = ?\)
- \(\log (7.42) = ?\)
- \(ln (ekt) = ?\)
- \(\log (ekt) = ?\)
- \(\text{antilog} \; 0.8704 = ?\)
- \(\text{antiln} \; 2.0042 = ?\)
- \(100.8704 = ?\)
- \(e2.0042 = ?\)
- 2.0042
- 0.8704
- \(kt\)
- \(kt \log e = 0.4343kt\)
- 7.42
- 7.42
- 7.42
- 7.42