Logarithm, power, index and exponent are all ways of expressing that same thing, and it's a very important concept indeed. They basically tell you how many times you multiply the number '1' by a particular number, and they are written as a small number to the top right of the 'base', which is the number that '1' is being multiplied by. In the diagram, 1 is being multiplied by a (the base), b times, to give c. Hopefully this will be made more clear in the following:
Let's take the number 4. If you say 41, you're really just saying you want 1 ? 4, just 4, just on it's own. Four. If you say it outloud, you would say 'four to the power of one' or 'four raised to the power of one', but you could just as easily say 'four' and it would still be true. As with any number, if it is 'to the power' of 1, then it will be 'itself', because you're just multiplying 1 by that number. It is raised to the power of 1, so it is just once. To put it generally, x1 = x.
Let's see what happens when we make it a bit more complicated. If we take 42, what we are saying this time is 1 multiplied by four twice - 1 multiplied by four multiplied by four, the product of 2 lots of four. 4 ? 4 or '16'. It's the same as saying 'four squared', because when you square a number, you simply multiply it by itself. If you look at a square, it's got two dimensions: length and width. In the same sense, four squared (or 42) has four along both 'dimensions', so you just take 4 ? 4 (? 1).
43 is four 'cubed' - 1 multiplied by the product of three 4's, or (1 ?) 4 ? 4 ? 4 (or '64'). Any number cubed is basically multiplying 1 by the number, then by that number again, then by that number again. You have three lots of that number. If it's raise to the power of 3, you multiply 1 together with 3 lots of that number. Speaking generally, if it's raised to the power of x, you take x lots of that number; so if you had 43932098 (which is unlikely, but possible), you'd take 3932098 lots of 4 and multiply them together.
After 3 you can't really show the dimensions, but the ultimate rule is still the same. The logarithm (or exponent or power or index) shows the number of times you multiply 1 by a number. 1 means that you just have to multiply 1 by the number once, 2 means you multiply 1 by the number twice, 3 means you multiply 3 times, and so on...
It becomes a bit more complicated if you have a negative logarithm. What does that mean? It can't mean you multiply something by itself a negative number of times - that just wouldn't work! Well, you're right, it's not that. This time, you divide by however many copies of the number you have. Going back to our example of '4', if you take 4 -1, it's the same as 1 divided by (1 lot of 4) or 1 divided by 4 or 1/4.
If we take 4 -2 it's the same as 1/42 or 1/16. Instead of multiplying 1 by however many copies of the number you have, you divide by however many copies of the number you have. x -a is the same as 1 divided by x a.
What happens, then, if you have something to the power of 0? Well, when you have a positive logarithm, you multiply 1 by the base that many times. When you have a negative logarithm, you divide 1 by the base that many times. When you have 0, you do neither - you don't multiply or divide 1, you just leave it as it is. You divide '0' times, you multiply '0' times, you do nothing, you take the 1. In that case, anything to the power of 0 = 1 (or generally, x0 = 1), and this is always true (except in the case of 0, because you can't multiply or divide by 0. You just don't get 0 raised to the power of anything).
If you don't understand that, don't carry on - it's very important you understand about powers before you can move on. Because now we get to the complicated bit....
What happens if you don't know the power? Sometimes you know the base, and you know the result, but you don't know how many times you've multiplied one by the base to get the result. If it's small, it's quite easy; if the base was 4 and the answer was 64, I know that the power is 3 (i.e. 43 = 64). However, if the base is 4 and the answer is 4,398,046,511,104, how do I know that the power is 21?
You do something called 'taking logarithms' or 'taking logs'. You rearrange the simple equation at the top of the page, to give you a slightly different form where the answer is the logarithm instead of the 'number'. So, if you know that ab = c, then you can say that b (which is the logarithm that you are looking for) can be found by taking the logarithm of c to the base a (i.e. to what power do you need to raise a in order to get the answer c?).
So ultimately 'logs' are just another way of expressing something, but this time they give the index / power / exponent / logarithm, which can be very handy with certain calculations.
One of the concepts which I always struggled with when learning this was 'taking logarithms' or 'taking logs'. When you want to keep an equation true, you have to do the same thing to both sides - so if you multiply the left hand side (LHS) by 5, you have to multiply the right hand side (RHS) by 5, if you take the square root of the LHS, you have to take the square root of the RHS. However, logarithms were a bit more complicated. They involved an unseen quantity, and I couldn't quite understand how it worked.
Eventually I got my head round it, because I realised what the logarithm was expressing. If something says 'logac = something' what it is saying is 'asomething = c'. Now, if you know that c = 5 (for example), then asomething = 5 as well as asomething = c. So logac = loga5. So, if you know that c = 5, then you also know that logac = loga5.
In this way, you can take the log of the LHS and the RHS in order to continue a calculation, because it will still keep both sides the same. If that's a bit too hard to get your head round, it's probably just as easy to say that sticking 'log' in front of something is a function, which you must do to both sides - so if you 'take the log' of the LHS (turning something such as c into something of the form logac) you need to take the log of the RHS as well.
Unfortunately logs are not always in the form you want them. However, there are certain well established (and mathematically proven) rules which, enable manipulation of a given logarithm into a more desirable form.
Imagine 2 separate logarithms (x and y) with the same base (a). If you multiply the two expressions together, you get the product of the results on one side (e.g. b?c), and the base to the power of the sum of the logarithms on the other (i.e. ax ? ay = ax+y).
The sum of the two logarithms (x+y) could be found by taking the logarithm of the product of the two results (b?c) to the base a (i.e. logabc). Since the two logarithms can already be expressed (i.e. x = logab and y = logac) then you could say that the sum of two logarithms with the same base is equal to the logarithm of their product, more commonly written as:
logab + logac = logabc
Imagine 2 separate logarithms (x and y) with the same base (a). If you divide one by the other, you get the division of the results on one side (e.g. b/c), and the base to the power of the difference of the logarithms on the other (i.e. ax / ay = ax - y).
The difference of the two logarithms (x - y) could be found by taking the logarithm of the division of the two results (b/c) to the base a (i.e. logab/c). Since the two logarithms can already be expressed (i.e. x = logab and y = logac) then you could say that the difference between two logarithms with the same base is equal to the logarithm of their division, more commonly written as:
logab - logac = logab/c
Imagine a logarithm. If you raise that expression to a particular power (e.g. to the power of m), you get the result raised to the power (e.g. of m) on one side, and the base to the power of the product of the logarithm and the new power (i.e. (ax)m = axm) on the other side.
The product of the logarithm and the power (e.g. mx) can by found by taking the logarithm of the result raised to the new power (e.g. bm) to the original base (i.e. logabm). Since the logarithm can already be expressed (i.e. x = logab) then you could say that the logarithm of a particular quantity raised to a particular power is equal to the logarithm of that quantity multiplied by that power, more commonly written as:
mlogab = loga(bm)
This one's pretty simple really. Because anything to the power of 1 is itself (i.e. if you multiply 1 by 'a' once, it will equal 'a') then when you rearrange to give the logarithm, the logarithm of any value, to the base of the same value, will be equal to 1, more commonly written as:
logaa = 1
This one's pretty simple again. Because anything (except 0) to the power of 0 is 1 (i.e. if you don't multiply or divide 1 by anything, you'll be left with 1) then when you rearrange to give the logarithm, the logarithm of 1, to the base of any value, will be equal to 0, more commonly written as:
loga1 = 0
This one's a bit more complicated, and requires you to have understood what has been said so far - particularly the concept of 'takings logs' and the logs law of powers.
This one's a bit more complicated, and requires you to have understood what has been said so far - particularly the concept of 'takings logs' and the logs law of powers. Imagine a particular expression involving a power: so ax = b. Take logs of both sides (the base this time is irrelevant, but for argument's sake we'll choose c). Using the logs law of powers we find that the equation simplifies to xlogca = logcb. If we divide both sides by logca then we get the logarithm (x) on the left hand side, and the division of two logarithms on the right. Notice that this is not the same as the logs law of subtraction as this divides the whole of the logarithm, not just what is 'inside' (i.e. one log is divided by the other log, not just the contents). Dividing the logarithm of b by the logarithm of a is not the same as taking the logarithm of (in this case) b/a.
Since the logarithm can already be expressed (i.e. x = logab) then you could say that the logarithm of a particular value (b) to the base of another value (a), can be expressed as the division of the logarithm to a third base of the original value (logcb), divided by the logarithm to the third base of the second value (logca), more commonly written as:
logab = logcb / logca
This may seem incredibly long winded and complicated, but it actually helps immensely. You'll notice that calculators don't generally allow you to choose the base - most will have a 'log' button, which (unless otherwise stated) means the base '10'. If the calculation doesn't involve the base 10 (though in science it often does) then you find youself in a bit of a pickle; however, you can use the logs law of changing base to change the base to 10. Excellent!