LOGARITHMS
Finding
the Unknown Power
for
Victoria
© 2003
Bart Alder
Creating a General Change of Base Formula
Logarithms were invented so that mathematicians could figure out problems involving the unknown power. In order to use logarithms, and to understand a little about how they work, you need to bear that in mind, always. Logarithms are for finding powers which you don’t know much about, they are the same thing as a mystery power. A power, an index, a logarithm... all wind up at the same mathematical place, they are all the same thing in the very end. That is THE key to understanding them, they are a mathematical tool for dissecting indices.
You do not need to say ‘power to the people,’ like John Lennon did. You can say instead, ‘logarithm to the people,’ because the logarithm always represents a power. Henry Kissinger claimed that ‘Power is the ultimate aphrodisiac,’ and we know that this can be redrafted, with tongue in cheek, to read ‘Logarithm is the ultimate aphrodisiac.’ No, it doesn’t seem too likely to me either but if logarithms and powers are the same thing...
A famous lord once wrote a famous letter to his son containing the words, ‘Power corrupts while absolute power corrupts absolutely.’ Which once again, we can translate as ‘Logarithms corrupt while absolute logarithms corrupt absolutely.’ We might not consider Austin Logarithms as so essentially groovy as Austin Powers, but we are assured that they must ultimately imply the same thing. Whatever you need to do to cement it into your head is worthwhile. Logarithms are just another fancy word for powers and that is exactly how we are able to investigate and find out about all their various properties. By remembering that logs are powers, you can investigate them for yourself and what you find out about them will hold true.
It’s not seriously worthwhile to sing ‘I’ve got the Logarithm!’ but it might help you to remember that logs and powers are the same thing in mathematics – emphasised to death only because this is such a very important thing to appreciate about logs. Whenever you see some equation such Log b a = L , tell yourself, that thing Loggy thing looks complicated enough, but all it really boils down to is some power, L, but it’s written in a loggy form which looks a little wooden at first.
Normally when we write some equation involving a power, like y = x3, we know what the power is, it is 3. That also means that the logarithm of this same expression is equal to three. We know that because logarithms are powers.
So that if we are given some value for x, (say x = 10) we can immediately calculate the value of y. We can do that calculation easily because we damned well know the value of the power (3), and we were also given some value for x. That is to say we have an equation where we start out by knowing everything, all the values, except for a single number, y and our job is to find this ‘y number’, having been handed the value of the power.
When we use powers, we are saying in effect, there is some number, x, which I must multiply by itself. The value of the power, obviously, tells us how many times we must multiply x by itself. In the example given, that means we take x = 10 and multiply it by itself three times to get y = 10 x 10 x10 = 1000 = x3.
So when we say a logarithm is a power, we can know immediately that it too represents the same basic idea of multiplying some number, x, by itself, some number of times, always. All properties of powers apply to logarithms. That is the moral of this particular story and we will see time and again how true that moral is.
But powers in most ordinary circumstances are already provided, as in the previous example where we were provided with the number three. Logarithms differ from the normal idea of powers in one very subtle way: When we say logarithm we mean to imply that we do not know as yet what the power is. We should, however, know everything else in the equation, including y. So suppose we are given y = 1000, x = 10 and we are asked to hunt down and find the power of ten which somehow ‘turns x into y’, we are looking for L where 1000 = 10L.
y = xL
Taking the Lth root of both sides of the above equation will also give,
L√y = x or y(1 ÷ L) = x
So that all problems involving logarithms really involve finding out what number a power MUST BE to make the equation hold true. We are given all the other numbers in some equation, but not L. As you can see, by finding L, we have found the power that turns x into y and by knowing the reciprocal of L, which is to say by knowing, 1 ÷ L , we have located the power which turns y into x. In both cases L is the same number. So the reciprocal of a logarithm is another logarithm.
Take for another example, the three equations below,
502 = 10L
502 = 8P
502 = 2I
Then we can say that there certainly exists some number, which we will call L, and this number L actually makes the first equation hold true. Also, we presume, there surely exists some power, P, which makes the second equation true. Lastly, we feel sure that there is some index, I, which will make the last equation be true and correct and safe enough to introduce to a mathematician.
That is to say we can find out what L must be by figuring out how many times we have to multiply 10 by itself, to make the number 502. So let’s begin there. How many times do we need to multiply ten by itself, to make the number 502. This problem will occupy us for the next section. The results we find will help us to understand logarithms very deeply, what logarithms achieve, how they work, what job they do, etc, etc, so please don’t give up.
We know that 102 = 100, which is obviously less than 502 and we know that 103 = 1000 which is larger than 502, and given that we are trying to solve 502 = 10L , we can be completely sure that the number L is greater than 2 and less than 3.
So we begin to get some idea about the value of L by writing that finding as an inequality. That idea of inequality means two things, both worth noting: Firstly that we can write a clear and exacting mathematical statement about the fact that we do not know something about the solution. We can qualify our lack of knowledge with such a statement. Secondly we can be very exact about our lack of knowledge. We can quantify it as well as qualify it, ie, we can be specific about what we do not know and so we are very precise about our degree of ignorance by specifying a definite range. That is what an inequality can sometimes represent, a range of possible values, only one of which is the answer we’re really looking for.
Now we do not yet know what L is as yet, but we’d sure like to know, and we can always use a trial and error system to gradually narrow our focus by homing in on the true value of L in various stages. That is what we are about to see, so please brace yourself for an argument about finding L through a series of better and better approximations.
2 <
L < 3 where 502 =
10L
You can even decide to invent a symbol to replace the word ‘where’, or you can use the symbol for ‘given’ from probability theory, which looks like this, | , so that we can eliminate all the English entirely and write a purely mathematical statement which means the same thing to a Japanese mathematician, as below.
2 < L < 3 | 502 = 10L.
This equation summarises what we know and also what we don’t know so far, which is a clever tool for mathematicians to have made. L is a power greater than 2 and less than 3, which also makes true the statement, 502 = 10L.
We can actually solve this directly by using standard logarithm tables or by using the log function on a calculator. We can calculate the expression,
Log 502 = L
And we will indeed get the right result. But before we just reach for the calculator, we should see how we might try to get closer and closer and closer to the answer by some thinking and reasoning. For instance, I can tell you straight away that L > 2.5 (L is bigger than two and a half). Let’s see why L, the power, must be greater than 2.5 and see why a fool like me can be so sure of it. That is 102.5 is going to be less than 502.
First we try estimating 102 . 5, just to see what the answer to this new problem will actually look like as a number, will 102 . 5 be larger or smaller than 502?
I can tell you the result will be too small by about 200, so my first guess (approximation) is that 102 . 5 = 300, but let’s see how it works so we can see how bad my guess was.
What is 102.5? Ie, how do you multiply ten by itself two and a half times? Suppose I say "solve 100.5", I am actually saying, ‘multiply 10 to itself less than once.’ In fact, I’m saying, ‘multiply ten to itself and do that self-multiplication exactly zero point five times.’ Eh? Can this be right? Yes! That’s exactly the problem we have run into. Partial self-multiplications!
What the $&@$#^% does a ‘half of a self-multiplication’ even mean, because it sounds like it either means something very stupid and wrong, or else it possibly means something very clever which is going to be far too theoretical to understand. Actually it means neither of those things. It will turn out to mean something you already know very well, which is quite nice to hear perhaps, because it means you don’t have to learn all that much which is new. Furthermore, it means that something you already understand turns out to imply partial multiplications, which is kinda weird. But it’s how it is.
1.1 Partial
Self-Multiplications of Base Ten
To explore this problem of "partial self multiplications" we will need to remember that our problem is a problem about powers. So we turn to look at our laws of indices from algebra to see if we can find out from a little thinking what a partial power would mean, if it had a meaning at all.
Note for example, that there is a law of indices which says that a(b+c) = ab x ac so we can indeed write,
102 . 5 = 10 (2 +
0.5) = 102 x 100.5
So that we can even isolate the partial multiplication and worry about it separately if we want to. This would seem to make things a little cleaner mathematically, but not any clearer intuitively.
That is until we consider only partial powers, ie, we see the law a(b+c) = ab x ac and we let b and c be partial powers. Then everything becomes clearer, not murkier, so please stay tuned.
As you might remember, whenever we have some number raised to a half power, we were actually talking about a square root. We are about to see how that must be true. That means 100.5 = √10. Now you say, "that’s great Bart, so you know some big law of algebra, you claim to know that a half power means a square root, but how can you be sure that really is what it means? Why does it mean that?"
Well I can be sure that is what it means because it is demanded by our laws for adding the powers, a(b+c) = ab x ac. You can check to see that this idea of a half power and a square root being the same thing absolutely must be true, by simply squaring both sides of the equation, 100.5 = √10. That is to say if a = b then we also know that a2 = b2.
You will then get,
(100.5 ) (100.5 ) = (√10) (√10) .
Where on the left side, a(b+c) = ab x ac says that you can now add the indices because the base 10 is the same. On the right hand side, you are merely multiplying two identical square roots... so we re-write the above to get the equation below,
(10(0 . 5 + 0 . 5) ) = 10
hence,
101 = 10
which was to be proved. So if you run the argument backwards, you can see there is no question that using a half power on the number ten, really is like taking a square root of ten. And so a square root really also is the same thing as partial ( ½ ) self-multiplication. In other words, when I said, ‘multiply ten by itself zero point five times,’ I could just as well have said ‘take the square root of ten,’ because those two statements mean exactly the same thing and this has now been proved.
This is why we can now confidently write any number in terms of its square, raised to a half power, knowing that this really is okay.
9 (0. 5) = 3 = √9
16 (0. 5) = 4 = √16
25 (0. 5) = 5 = √25
36 (0. 5) = 6 = √36
49 (0. 5) = 7 = √49
64 (0. 5) = 8 = √64
(n2) (0. 5) = n = √n2
And so on for any value of n.
Now we should recall that we were interested in evaluating 10(2+ 0.5) = y, to see how the number y compares to the number 502. We can now begin to do that evaluation because we know what a half power means, it means we take the square root of something. You might remember that we recently wrote down,
102 . 5 = 10 (2 +
0.5) = 102 x 100.5
Well we can figure this bolded part out without too many worries. We know 102 = 100 and we also just found out that 100.5 = √10. So we can re-write the above equation as,
102 . 5 = 10 (2 +
0.5) = 100 x √10
But √10 is going to be very close in value to √9 = 3 and so this was the source of my estimate of 300. I replaced 10 with 9 and took the square root of the easier number to get my first approximation. So I did this,
102 . 5 » 100 x √9
» 100 x 3 » 300
In other words, I cheated in order to get an approximation.
I know that √10 and √9 will be quite close in value so that the error is kept small. If you evaluate √10 to get a truly accurate approximation, you find 100.5 = 3.16227766...
That is to say,
100 x √10 = 100 x 3.16227766 = 316.227766
So if we use two and a half as a power, and ten as the base, this is the result we find and we have even seen why this is the result we find. It was also why I could guesstimate that two raised to a power of 2.5 would give roughly three hundred as a result. My error is a little over 5%, which is not at all so bad for an estimate. Either way, we were hoping to find the solution to 502 = 10L and evaluate L. We now know that L > 2.5.
So we have to now write our inequality statement and include this new piece of extra precision. Where before we said,
2 <
L < 3 where 502 =
10L
Now we have decreased our ignorance about L and we write,
2.5 < L <
3 where 502 = 10L
Clearly we are gradually homing in on the true value of L by a process of making better and better approximations by using partial powers and what they come to mean mathematically (ie with logical consistency). Next we should try the index of 2.75 to see how we go by using that value. I’d like to show you what a power of 0.75 comes to mean, if only because it’s so very fundamental to understanding logarithms and powers and their behaviour. It ultimately means a three quarter power and you can deduce that because you already know that 0.75 = ¾ .
That is, if we are considering a three quarter power of ten, 10¾ = j, to evaluate j, we first cube ten, to get a thousand and then we take the fourth root of a thousand (the fourth root of a thousand means we take the square root, of the square root of 1000.). Ie, we know that two successive square roots is the same as a fourth root. √(√1000) = 4√1000 and so we have,
4√1000
= 4√(10 x 10 x 10) = (103)¼ = 10¾ = 100.75
You can check on your calculator that these are all the same result. You can also show that we could have written it out another way. A fractional power such as ¾ can mean we first cube ten and then next take the fourth root, or else it can mean the total inverse. It can mean that we first take the fourth root of ten and then next cube that result. It makes no possible different to the answer according to our law of powers. So we could just as well have written out,
(4√10)3 = (10¼)3 = 10¾
= 100.75
Rather than walk you through a second bunch of estimates for calculating ten raised to the power of 2.75, I’ll just cheat with my calculator and state its answer.
10(2.75) = 562.3413...
So we are definitely getting very close now because we want to get 502 as an answer for 10L and here we have some number a little bit bigger than 562. So once again we write our inequality statement out, correcting it a little more...
2.5 <
L < 2.75
where 502 = 10L
We note that L will only be a little bit less than 2.75 because 562 and 502 are reasonably close. Suppose we try L = 2.7, then we shall find that,
10L = 10(2.7) = 501.1872336
which is very close to 502, painfully close, but still not quite there... so again we write our inequality and again we think to include this new correction...
2.700 < L <
2.750 where 502 = 10L
And you can see that we are indeed homing in on our true value for L.
Now if you punch the number 502 into your calculator and hit the Log button,
you should see the following number appear, 2.700703717...
This is our answer and you can see how fast the answer came out compared to the method of homing in on a value for L by better and better approximations.
This convoluted idea of homing in on good estimates for fractional powers of the base ten is exactly the same job as is done by base ten logarithms. You can see how much time is potentially saved by having a Log button on your calculator, or by owning a set of Log tables. The effort we had to go into, to get some estimate of the number L was somewhat excruciating compared to the time saving result of hitting a single Log button on a calculator. So 102.700703717 = 502. And that is comforting because it saves effort and time.
Hence, you can indeed say that,
L = Log 502 = 2.700703717...
you will also see that same thing written as,
L = Log10 502 = 2.700703717...
where in the second case, all we did was add to our equation the fact that the base we used, was base 10 and we wrote that fact as a subscript.
Suppose I had not written 502 = 10L, but instead I had written 5020 = 10S. What would this mean? I can tell you immediately without a single calculation that just because we know L = 2.700703717.... that S will be equal to S = 3.700703717... Can you see why the digits are all the same, except the first? Remember that we found,
502 = 102 x
10(0.700703717...)
Well what will 5020 be? It is equal to 502 times ten.
5020 = 101 x
102 x 10(0.700703717...)
or,
5020 = 103 x
10(0.700703717...)
Since all we have to do to get 5020 from 502 is multiply 502 by a factor of ten. That just means we add the number one to the index because ten was our base. Thus we can immediately know,
502 = 102
x 10(0.700703717...)
5,020 = 103
x 10(0.700703717...)
50,200 = 104
x 10(0.700703717...)
502,000 = 105
x 10(0.700703717...)
And so on, ad infinitum. Each row is just ten times larger than the row above it. We can therefore add the number one to the power each time we go down one row. Or we can go the other way... we can subtract the number one from the power each time and divide by ten, to create decimals, as below,
502 = 102
x 10(0.700703717...)
50.2 =
101 x 10(0.700703717...)
5.02 =
100 x 10(0.700703717...)
0.502 = 10-1
x 10(0.700703717...)
0.0502 = 10-2
x 10(0.700703717...)
By working out one example, we have already worked out an infinite number of similar examples where only factors of ten are involved. That was done by knowing that logarithms are essentially the same thing as powers, and by assuming as true the idea that fractional powers behave according to the same laws of algebra as do the integer powers.
Now we started all of this by wondering how we could solve 502 = 10L for some number which we called L. We said that this was exactly the same problem as,
Problem 1.a: Evaluate, Log10 502.
Problem 1.b: Solve for L, 502
= 10L
These two different looking things are actually the same damned problem, the first is expressed as a logarithm of base ten, the other is exactly the same idea, but we stated it in terms which might seem a little bit more familiar. So we now know how to use our calculator to solve any problem in either form, where we have base ten and an unknown power, L. We punch 502 into the calculator and hit the ‘Log’ button. That’s it! So it’s very simple for base ten because our calculators always use logs to base ten. That is what the Log button does. It evaluates the unknown power. To summarise, whenever you have the problem of a power which you are asked to find, and whenever your base is in the base of ten and you have some mystery power, you are being asked to use the Log button on your calculator to find that mystery power.
Problem 2.a: Evaluate, L = Log 6117.
This is clearly the same thing as,
Problem 2.b: Solve for L, 6117 = 10L.
If you type the number 6117 into your calculator and then hit the Log button, you should get the result, 3.786538481... as your answer. You have found that L = 3.7865... and you have also found that Log 6117 = 3.7865... because they really are the same thing!
We can even write,
6117 = 10 (Log 6117) = 10 L
That openly screams at you that L = Log 6117 really is a power no matter if you call it "L" or if you call it "Log 6117", it’s a power all the same. All loggy things are powers, no qualifications. No correspondence will be entered into... :))
Logarithms are always unknown powers requiring your evaluation. The logarithm button on a calculator finds unknown powers of the base of ten only so that all the problems given so far, rely on that fact. This is extremely important to appreciate. That the Log button handles only base ten is a fact of calculators, not mathematics. If we, as humans had a different standard base, say base 16, our Log button on our calculators would only solve problems like,
16L = 256
But this is not the case. We use base ten so instead (for the time being) we are completely stuck with problems in the form of,
10L = 256
Where ten is our preferred base. In our long-winded example, we used the base ten for that very reason, so that we could examine the "simplest" case first before seeing how other cases involving some other base, say base nine, will be related to the solution we just saw. So our decision to use base ten was not entirely arbitrary, it was based on the properties of human calculators.
In any sense, irrespective of your base, the method I just showed you of making better and better estimates of the unknown power by trial and error, will always work. And whenever you use logarithms, you are deliberately avoiding this long-winded style of finding the solution to the unknown power by trial and error.
So logarithms might seem painful to learn, but actually they save masses of time in situations where you already know everything but the power because they stop you from trialling and erroring your way into old age. Logs are an avoidance scheme for a bigger effort!
Below are a set of problems of base ten which you can evaluate by simply using your calculator’s Log button. The answers are given at the end of the essay. I have set the problems in both formats to hopefully give some practice at translating between the two systems of notation. Just remember that every time you are being asked to find an unknown power, you are really being asked to solve some kind of problem involving logarithms. That makes it very easy to know when to use Logs. You use them whenever and wherever you have a power which you do not know the value of. No exceptions.
Problem Set A)
Problem 1) Solve for L, 8 = 10L
Problem 2) Evaluate, Log10 2649
Problem 3) Solve for x, 928 =
10x
Problem 4) Evaluate, Log10 359
Problem 5) Solve for p, 3589 =
10p
Problem 6) Evaluate, Log10 2588
Problem 7) Solve for g, 4563
= 10g
Problem 8) Evaluate, Log10 25
Problem 9) Solve for L, 2123 =
10L
Problem 10) Evaluate, Log10 659875
We now move away from base ten to generalise the idea of the logarithm. After all, we want to solve problems like,
Find P, given, 502 = 8P
Where now we are using base eight and not the ‘easiest’ base of ten. Let’s see how we can do it using what we already know...
2.0 Logarithms
to Any Old Base
When we wrote the equation 502 = 10L we found that L = 2.700703717...
Our choice to use 10 as a base for all the standard logarithm tables was completely arbitrary.
When humans decided to stick to a single standard base, it was likely decided on the number of digits that are found on two human hands, so that when the tables of Logarithms were produced, they were all written for powers of ten only. The calculators we all use are set up the same way, with the same arbitrary convention. When you see the Log button on a calculator, you use it knowing that it solves for base ten and only base ten. Same with the Log tables you find printed in the back of all of the older maths textbooks. These wonderful tables are for base ten only.
You might say, "well that’s all but useless because we can have any base other than ten and our precious Log tables become useless!" This turns out to be untrue. We can use our Log tables and our calculators to evaluate an unknown power to any base at all. We just need to translate the problem into base ten. Then we can easily solve it. So all that follows is really just a process of changing all problems involving unknown powers into the base of ten.
So suppose we have the following problem...
502 = 8P
We know already from working out the last problem of 502 = 10L that to find L we can just plug the number 502 into our calculator and then hit the Log button. We found from this small effort that L was equal to 2. 700703717...
Well, we know that we have a different base here, so that we just cannot solve 502 = 8P by typing in 502 and then hitting the Log button on our calculator because humans do not use base eight, we use base ten, and assuming that the base is always going to be ten gave us the value of 2.700703717... as the power, and this was meant to imply that,
502 = 10(2.700703717...)
This time we still want to find a result of 502, but we no longer have a base of ten, but instead a base of eight. That is to say we have to figure out, now, how the heck we can find P when our calculator only does base ten?
Well, we know that we can in fact always resort to the method of writing out inequalities, the long-winded method which we saw was accurate but slow and painful. That looks too painful however, and our mathematical intuition perhaps tells us that there might well be a way to convert our entire set of log tables from base ten to base eight through a single, simple system which always works. So what the hell do we do to make this conversion work out okay?
Well if you have done the problems in problem set A) then you have in fact already partly solved this problem of raising (unknown or mysterious) powers to a base other than ten. Problem 1 in that problem set, was expressed as,
Problem A.1: Evaluate 8 =
10L
That is to say that we can, if we want, express the base of 8 as the number 10 raised to some unique and distinct power. That was our ‘former’ finding. Any number, x, can be expressed as some other number b raised to some power L.
And to find the value of L we take Log b x = L, which is to say for b = base = 10 that we can simply plug x into our calculator and then hit Log. The result is L.
So we solve that part of our problem now, we have 8 = 10P and we
need to know P so that to find P we need to evaluate Log10 8 which
is indeed in base ten, so we can put 8 into the calculator and hit the Log
button and we get, Log10 8 = 0.903089987...
That means we can now write, 10(0.903089987) = 8 and feel fully justified in that move. Let’s look again at the problem we were given to solve, which was 502 = 8P.
We should always write our various equations out side by side so that we know exactly what it is that we are being asked to solve;
a) 10(0.903089987) =
(8)
b) 502 = (8)P
Where we also know from previous workings out that,
c) 502 = 10(2.700703717)
So we write all three equations out together in a list and just stare at them.
a) 10(0.903089987) =
(8)
b) 502 = (8)P
c) 502 = 10(2.700703717)
We now have everything we need to compute P. That is to say, we can now figure out the logarithm (power) to a base other than ten, ie we can write Log laws for an arbitrary base of eight. We could have chosen some other base and the same laws would look very similar. But we chose 8 as a base, so let’s keep going to see how it all works. We can worry about other bases later on when we have understood this particular problem.
Firstly we should notice that equations a) and b) both contain the number 8. Next we should note that equations b) and c) both equal the same value of 502. That is to say b) and c) are equal to 502 and they are therefore equal to each other. We combine b) and c) into one line, to derive,
502 =
(8)P = 10(2.700703717)
Next we can see that because a) and b) both feature the number 8, we can do something interesting with them too. Since equation a) assures us that 10(0.903089987) = 8, we can find the number 8 and ALWAYS substitute the number 10(0.903089987) because really they are the same value in the end. But we just saw that,
502 =
(8)P = 10(2.700703717)
And I just said that where we see the number 8, we can substitute 10(0.903089987) because they are the same number. Well above we have the number 8, so we definitely do this substitution. And again, we derive,
502 =
(10(0.903089987))P
= 10(2.700703717)
And we are very nearly finished because we now have the same base (base ten) on both sides of the equals sign. This is an important step to have taken, as we shall soon see.
Now we need to remember another, second, law of algebra that says (ab)P = (aP)b = abP , well here, because we converted all our numbers to base ten rather than base eight, we have a = 10, so we can say, (10b)P = 10bP .
Once again, here we have b = 0.903089987. So we can continue.
(100.903089987)P = 10 (0.903089987 ) P .
In plain English, whenever you raise a power to another power, you can always multiply the two powers involved. In our equation above, we see that we have the power 0.903089987 raised to the power of P. So because we can multiply the two different powers, P and 0.903089987... we can move P inside the brackets and derive,
502 =
(10(0.903089987) P ) =
10(2.700703717)
Where we have created the same base of ten on both sides of the equation. This is an important step, so try to remember it. Why is it important? This same base on both sides of the equals sign can only mean that the indices are also equal. If 10a = V = 10b then we can be sure a = b because the bases are the same and the results are the same, both are equal to the number V.
If this were not so, then the same number, ten, raised to two different powers could equal the same result, which is clearly illogical. Hence, because we made the base ten the same on both sides of this equation, we can actually lose the base entirely because we have arranged it so that we definitely made the bases the same. So we drop the bases and look now only at the indices and make a mathematical statement about our finding; if the bases are the same, and the two numbers both equal the same number, 502, then the two indices are also the same, thus, we can drop the bases and consider only the indices! A bold move indeed.
(0.903089987) P = 2.700703717
Which we can solve for P by dividing both sides by (0.903089987)...
P = (2.700703717) ÷ (0.903089987)
So
P = 2.990514518
In other words we have solved our original problem,
502 = 8P = 8(2.990514518)
Which you can check for yourself using your calculator by using the yx button.
Raise 8 to the power of 2.990514518 and you do indeed get 502. To use a calculator to compute this, you first punch in the base number, so you will type the number 8. Next hit the yx button which raises eight to some power and then type in the power itself, which is to say you type in 2.990514518. Lastly, hit the equals button. You should see 502 as the result. You have just raised 8 to the power of 2.990514518. And you found 502 was indeed the result. So we have definitely found a way to find out which power we need to raise 8 to, to get 502 as the result. Given 8P = 502, we have deduced P by using not base eight, but instead base ten. A very tidy trick indeed!
We have now used base ten logarithms to find logarithms in base eight. We ask ourselves, how did we do it? Well, we got interested in the factor, Log 10 8 = x, which is to say we got interested in solving the problem of the unknown power, 8 = 10x and then when we knew how to make ten from eight raised to some power, we did our substitutions so that all the numbers in our equations were in the same base of ten and not the base of eight. That allowed us to drop our bases and consider only the indices.
After getting rid of all bases, by making all sides of the equation in the same base, we wrote a statement out about the indices alone and lastly we solved our derived equation of indices to locate a value for P. If you look at the equations above, you will see an equation of the form P = (2.7007...) ÷ (0.90308...). We are now interested in this form because it actually contains a lot of general information about how all logarithm tables of all different bases are related from one base to the next.
This is a very general idea.
It means that one set of logarithm tables contains exactly the same data as a set of Log tables to a different base. In other words, if humans used base 18 instead of base 10, our logarithm tables would look different, but the kind of information contained in the two different tables would be exactly the same.
Let’s see why.
2.1 Creating
a General Change of Base Formula
If we care to look for a general method for doing this to any base, we will see that this form of equation, P = (2.7007...) ÷ (0.90308...), contains a lot of useful information, all the information we shall need! That is to say, one table of logs written in base ten is indeed very much like another set of logarithms written for base eight. All that makes the two sets of tables differ is a single factor. If we estimated something other than 502, we would still find the same factor separating base ten and base eight. The factor of (0.90308...).
Let’s see that this is true. We need to remember where all these numbers actually came from originally. That will give us a strong clue as to how to write Log tables to some other base than ten.
P = (2.700703717) ÷ (0.903089987)
Is the equation we have to now work on, by going backwards through our own work and making the correct substitutions. We said 502 = 8P so that this is where P comes from and we can re-write this as,
P = Log 8 502
Where our subscript of eight tells us we are using logarithms, but not in our old base of ten. So we substitute that into P = (2.700703717) ÷ (0.903089987) and we get,
Log 8 502 = (2.700703717) ÷ (0.903089987)
Next we consider the term 2.700703717. Where did this come from?
Well, this came from evaluating the expression, 502 = 10L so that we rewrite this equation as a logarithm instead to get,
L = Log10 502 = 2.700703717
So we substitute that into our P equation once again, and we find out that,
Log 8 502 = (Log10 502) ÷ (0.903089987)
Lastly, we want to know, where did (0.903089987) come from? If we think back we remember that this came from finding out what power we needed to raise ten to, in order to get the result of eight. Ie, we evaluated Log 10 8 = (0.903089987). So lastly we substitute that into our P equation. We get,
Log 8 502 = (Log10 502) ÷ (Log 10 8)
This is a change of base formula for bases ten and eight, which we can further generalise by noticing that ALL the numbers we used were actually also arbitrary.
Suppose I say now, "I don’t care much for the number 502. I want to instead use the number 6009. I want to know what number I must raise 8 to to get 6009." Then we can just take the above equation and replace every instance of 502 with 6009 to get,
8P = 6009
or
P = Log 8 6009 = (Log10 6009) ÷ (Log 10 8)
Which is,
Log 8 6009 = (Log10 6009) ÷ (0.903089987)
Suppose I say now, I didn’t really want 6009 either, but I wanted instead to use 989, which is how old I feel right now. Then I can indeed write, 8P = 989, or,
P = Log 8 989 = (Log10 989) ÷ (Log 10 8)
Or,
Log 8 989 = (Log10
989) ÷ (0.903089987)
Suppose now instead, I say, "I wasn’t really interested in the number 989, but some number r, which I cannot exactly remember the value of." Then we replace 989 with the number r instead and we have 8P = r, or,
P = Log 8 r = (Log10 r) ÷ (Log 10 8)
And the equation will always hold just as true.
Now here’s a very clever step. I say "Did I say base eight? Damn! Sorry! I meant to say base 7 instead..." then we can simply find all instances of 8 in the equation above and replace it with the number seven. We then get 7P = r, or,
P = Log 8 r = (Log10 r) ÷ (Log 10 8)
Becomes,
P = Log 7 r = (Log10 r) ÷ (Log 10 7)
And the equation still holds just as true. Clearly if I said... "Did I say base seven? I really meant base b, but I cannot remember the value of b right now." Then we replace the number 7 with the term b instead and we derive,
P = Log b r = (Log10 r) ÷ (Log 10 b)
And finally, if I now add, "Hang on a minute, we used base ten to do all of our calculations and I wanted that to be in base twelve. Sorry about that." You can say "No sweat, loser." And write,
P = Log b r = (Log10 r) ÷ (Log 10 b)
Becomes,
P = Log b r = (Log12 r) ÷ (Log 12 b)
And if I said "Wait, I didn’t want base twelve at all, I wanted some base called a, and I forget what the number a is!" You reply, now angry. "I can do it, but I’m beginning to hate you." As you write,
P = Log b r = (Log a r) ÷ (Log a b)
And you have derived the law of changing bases of the logarithms from a process of assuming generalisations were possible, which they are in fact. It is clearly a law of logs which enables you to change from one base to another. You have not said what either bases, a or b are, in real values, you have not even said what the result, r, should be. Your formula will work all the same.
That is also to say that every table of logarithms is very like every other table of logarithms. They contain exactly the same key information, even though one table will look different to another, the data in any table is resolute and defiant and it will not go away under a change of base, it will merely mean that all values in the table are changed by a single factor of conversion. That means that if I give you a table of logarithms in base ten and say to you, "convert all the numbers in this table from base ten to base eight and create for me a new table of logarithms for base eight, you would say let a = 10, let b = 8 and we would be back to using,
P = Log 8 r = (Log10
r) ÷ (Log 10 8)
Which says that the way to make your table is to compute (Log 10 8) once, and then use it on every value in the base ten log tables to derive the equivalent value to write your base eight log tables. You need not concern yourself with knowing ten thousand different conversion factors to transcribe tables from one base into a completely different base, just one conversion factor will do, always. All you need to know is (Log 10 8) and to have the set of logarithm tables for base ten and you can happily and mindlessly compute your entire set of logarithm tables for base 8 by knowing only one conversion number.
We know that is true of bases other than eight and ten as well because we have already seen that you can always translate the values found for log tables in base a to a set of tables in base b by interesting yourself in the single factor, Log a b.
It makes no difference which base you use for your general set of log tables, say you use the conventional base ten tables. Your tables in base ten will behave the same way as some other person who uses base one million and thirteen, in that each number in your table will differ from the corresponding number in their table by the same single factor, irrespective of which part of the table you are considering.
Let’s go through a few more examples converting from base ten to some other base to get used to the idea a little.
Example Un
We know for example that if we are given 1000 = 10L then L is equal to three. We write this as, Log10 1000 = 3.
Now we wish to evaluate the problem 2P = 1000 and solve it for P. Our base is 2 and not 10, so we know we have to ‘convert our tables’.
We write out all of our equations,
Log10 1000 = 3
Log2 1000 = P
And lastly we will need to know the answer to the problem 2 = 10n, which we write as,
Log10 2 = n
We can evaluate n by putting the number 2 into our calculator and pressing the Log button because it is a problem posed in base ten already. We find, 10n = 2
Log10 2 = n = 0.301029996, or 2 = 100.301029996.
Remember that we had, P = Log b r = (Log a r) ÷ (Log a b)
We have then r = 1000, b = 2 and a = 10. Substitution gives,
P = Log2 1000 = Log10 1000 ÷ Log10 2
Which becomes,
P = Log2 1000 = 3 ÷
0.301029996
Which when we evaluate it, becomes, P = 9.965784285
In other words, 29.965784285 = 1000.
Example Deux
Suppose we are asked now to evaluate 16P = 91.
Former considerations tell us that the values we are interested in are going to be Log10 16 and also Log10 91. By knowing those values we can figure out P by converting it into a problem of base ten.
Log10 16 = 1.204119983 or 101.204119983 = 16
Log10 91 = 1.959041392 or 101.959041392 = 91
If 16P = 91, then we now know 10P(1.204119983) = 10(1.959041392)
So we can drop the identical bases and derive an equation about the indices,
P(1.204119983) = (1.959041392)
Solving for P we get,
P = (1.959041392) ÷ (1.204119983) = 1.62694866
Which means that 161.62694866 = 91.
Example Trois
Suppose we are asked now to evaluate 3000P = 6,000,000.
We are now most interested in Log10 3000 and also Log10 6,000,000.
Log10 3000 = 3.477121255
Log10 6,000,000 = 6.77815125
We now know 10P(3.477121255) = 10(6.77815125)
So we can drop the bases once again and derive,
P(3.477121255) = (6.77815125)
Solving for P we get,
P = (6.77815125) ÷ (3.477121255) = 1.94935717
Which means that 30001.94935717 = 6,000,000.
Here’s another set of ten problems to get to familiarise yourself with the proceedures involved in converting from some base x to base ten.
Problem Set B)
Problem 1) Solve for n, 7n = 6009
Problem 2) Evaluate, Log 2 28
Problem 3) Solve for c, 1970 = 33c
Problem 4) Evaluate, Log 39 1964
Problem 5) Solve for f, 1943 = 28f
Problem 6) Evaluate, Log 18 1985
Problem 7) Solve for g, 243 = 9g
Problem 8) Evaluate, Log11 125
Problem 9) Solve for φ,
2127 = 15φ
Problem 10) Evaluate, Log13 256
Problem Set C)
Now convert all of the problems above, so that if the problem is stated in the form ab = c your solution should be put in the form b = Log a c. And conversely if the problem is stated in the form Log a c then you will need to write this as ab = c. You will find that you need to create a symbol b, in this second case because I wrote the problems in the form, evaluate Log a c, and did not say instead, evaluate b = Log a c. So if you are given,
Evaluate, Log 19 34. You might answer this by saying, If b = Log 19
34, then 19b = 34
3.0 A
Physical Use for Logs (From A Mathematician’s Delight, by WW Sawyer )
The logarithm, says Sawyer, is like a rope being coiled around a post. You may know that if you turn a rope around a log a few times, the rope becomes very hard to pull because of friction. One turn and you can still pull the rope. But friction multiplies with every turn by a constant factor. So, suppose you have a frictional measure of the rope against some arbitrary log / pole / object and that one turn gives you a measure of ten in certain units. Suppose ten is the number of grammes you can hang from the rope before it begins to slide against friction.
Now after two turns you find that you can hang much, much more than twenty grammes before the rope begins to slide. You can hang a hundred. With three turns you find you have reached a thousand grammes and so one kilogram is required before the rope gives into friction and will slide. With four turns you are at ten kilos so you know that five turns of this rope will support the weight of a 99 kg person with some precariousness. Six turns will put beyond doubt the issue of safety and so you know from turning the rope six times, it will support the weight of any individual.
This remarkable fact of weights and friction around a uniform surface with a uniform rope, is like saying that each time you turn the rope once around the pole, you are adding one to the power of the weight it will hold. Friction produced by coils of rope, mimics the behaviour of powers very closely because the frictional effect multiplies with every turn.
Now we started in units of ten grammes, and this ‘represents’ the case where we’re using base ten.
Logarithms, says Sawyer, are exactly like the number of turnings of a rope around a pole. So that a partial turning of the rope 3.06.... times around the pole, is like 3.06... as a power to some base, which we selected to be ten. That is to say that you can think of the number 8190. 089 as a pole and rope arrangement which can hang eight grammes, say, but where the rope turns around the pole 190.089 times. As we saw before, 0.089 as a power means taking a fractional root of the base, in this case eight.
So there’s a kind of visual game of imagination which tells you how logarithms behave as though they are also ropes turned around poles. Any way you look at them, they are the same kind of thing. They represent the idea of bases and powers sharing a kind of eternal relationship that you can infer things about, you can write rules about this relationship and those rules are the famous laws of logarithms.
This is why our problem of logs which we originally stated as;
Problem 58789345.d Evaluate
p = Log (14) 5107
Is entirely equivalent to solving the physics problem, if I have a weight K, of 5107 grammes, and I have a rope and a pole, and one turn around the pole of the rope can safely suspend a weight of 14 grammes without the rope slipping, then how many turns, p, will I need to safely suspend the weight K?
You will find in physics that logarithms are used frequently because there are a lot of cases where we find that in order to solve some equation we are forced to deal with the circumstances where we know the value of everything but the power and we are then forced to use logarithms as an aid to rapid calculation. But more than that, logarithms are not simply human tables... to some extent they are because we write all our numerals in base ten. But we should not ever fool ourselves into thinking that our numerals have anything at all to do with the patterns we have found. The laws of logarithms are surprising and as we shall see, they lead to a very astonishing part of the mathematical garden precisely because no matter what your base, once you begin using logarithms a natural base appears which simply glares at you, and after too long of ignoring this natural base you’re forced to look into it and what do you find but the base of god.
4.0 Natural, or Napier Logarithms and Their Origins
Now you can use the laws of logarithms to take logarithm tables written in one base, say base ten, like our tables are, and you can convert them all to logarithms in another base very easily. There’s a law to handle that, a change of base law, which we saw.
And another interesting thing is that no matter how you assemble your logarithm tables, which base you use that is, you always find that you have to add small corrections as you go. You can calculate each logarithm rather well in any base, but the corrections that come along, well... the closer and closer you get to calculating smaller and smaller powers of that base, the more clear it is that the correction you’re getting is a very definite kind of correction. What that means will become completely clear.
Suppose we are considering a small power of ten. This must be done if we are to compute tables of logarithms from basic principles.
We need to know small powers so that we can do our trick with the fractional indices and evaluate any complicated fractional power by evaluating good estimates of small fractional powers and using them to deduce what the value of a complicated fractional power should be.
So we should consider,
100.1
100.01
100.001
100.0001
100.00001
... because they will assist us in evaluating something like 100.938757. How will they assist that? Well observe the following. We are about to produce a long hand estimate of 100.938757 by using the values above.
100.9000000 = (100.1) 9
100.0300000 = (100.01) 3
100.0080000 = (100.001) 8
100.0007000 = (100.0001) 7
100.0000500 = (100.00001)5
100.0000070 = (100.000001) 7
Hence we are entitled to write,
100.938757 = (100.1) 9 x (100.01) 3 x (100.001) 8 x (100.0001) 7 x (100.00001)5 x (100.000001) 7
So by knowing small values of the powers we can make any decimal number into a power and get some estimate of the value. This is actually how the first logarithm tables were constructed, from knowing very small, fractional powers of base two, not base ten, but the idea is exactly the same.
Notice that in this situation we were only discussing powers which were very small, nearly zero but not quite zero. Well, we know that anything to the power of zero is equal to 1. So ten to the power of zero point something will equal something a little bigger than 1. This idea of ten to the power of zero point something we can write as 10k and the result will be greater than one by some small amount, which we will also write in terms of k, the power, and some other number, n. Why we do it this way will become clear, so hang in there. Remember, k is a small number close to zero and we then write out,
10k = 1+ kn
So k times n is a part of an additive correction to the result of 100 = [1 + 0(n)] where k=0. k appears on both sides of the equation. It is the power we’re concerned with calculating out on the left side, and here on the right side we also have (1+ kn), a term which is NOT A POWER.
We want k >0 now, but we also know that k is a very small number, it’s only a little more than zero. Either way, we are about to use 10k = 1+ kn.
Let’s rewrite the equation, solve it for n by subtracting 1 from both sides and then dividing both sides by k. We find,
( 10k - 1 ) ÷ k = n
Now we will make a small k ¹ 0 and see what we get for n, from using a few different values of k.
For k = 0.1, (100.1 - 1) ÷ 0.1 = n = 2.589254118
For k = 0.01 (100.01 - 1) ÷ 0.01 = n = 2.329299228
For k = 0.001, (1000.1 - 1) ÷ 0.001 = n = 2.30523808...
For k = 0.0001 , (100.0001 - 1) ÷ 0.0001 = n = 2.3028502...
For k = 0.00001, (100.00001 - 1) ÷ 0.00001 = n = 2.302612...
For k = 0.000001, (100.000001 - 1) ÷ 0.000001 = n = 2.30259...
We find in fact that as k continues to get smaller and smaller, n = 2.302585093... or, substituting,
10k = [1 + k (2.30258...)]
The whole point is this, when we let k get smaller and smaller, we find that the number n approaches a constant number 2.30258... to more and more decimal places! Even as we change k, make it smaller and closer to zero, n remains something of a constant and the number of decimal places it is constant to, only ever increases as k shrinks towards zero.
Now, if you used base 9 to compute your tables the corrective factor you discover around the smallest powers is not 9k = [1 + k (2.30258...)] but instead, it’s,
9k = [1 + k(2.197224...)]
And if you’d used base three, the corrective factor of n would be more like,
3k = [1+ k(1.0986122...)].
So this corrective factor n, gets close to n=1 somewhere around base 3 and n definitely changes according to the base you decide to use. Note that when the corrective factor really is 1 + k(1.00000000), we have a rather interesting kind of base as well. We have a base which has no extra corrective factors involved in its behaviour and our original equation bk = 1+k exactly, where at the risk of drilling it in too far, k was a small power a little more than zero. The term b represents a base which we have yet to provide and n has vanished from the equation.
There is, therefore a simplest possible base, b, in which certain essential calculations are made especially easy because n=1 and this is a base which nature also respects. We have to find this number e, because in no way can it remain hidden for too long.
b = e = 2.718281828459045235360287...
So that for very small powers, k, we can write ek = 1+ k and the corrective factor of n has completely disappeared. We can write this as a statement of logarithms too.
Loge [1+ k] = k. (for small k)
Or,
Log 2.718281828... [1+ k] = k. (for small k)
Whereas for other bases, not equal to e, we would need to include the factor n as well. Since we’ll use it later, I will call our rule, Rule A)
Rule A) Log a [1+ kn] = k. (for small k)
Now very reasonably you might yell. ‘2.718281828... IS SUPPOSED TO BE NATURAL?’ and you might even scream, ‘AND YOU THINK GOD CHOSE THIS? HAVE YOU LOST ALL YOUR LITHIUM TABLETS?’
To which I, personally, would answer ‘yes’ and ‘maybe’ and ‘definitely’, respectively.
Now let’s see something really remarkable. We noted that when we used base ten we were compelled to use a correction which included the corrective term n = 2.3025... well now we try to raise e to that power, the power of the corrective factor, n, ie we consider not ek, where k is small, but en. The natural base raised to the corrective factor. You will get for n = 2.30258... the result that e2.30258 = 10.
NOW HANG ON A COTTON-PICKIN’ MOMENT!
10 was also the base we used when we found this corrective factor! Now we raise a completely different base to the corrective factor and we somehow magically get back to the number 10 – which is certainly very darned strange!
When we used base nine we found we got a different value for n. We got n = 2.197224... Now try raising e to the power of 2.197224... I dare you. Nine was the base we were computing in this time. And sure enough, just like magic, e 2.197224 = 9.
When we ‘computed’ log tables in base three we got a resulting corrective factor of 1.0986122 and this is nothing other than log e 3 = 1.06986122. Or e1.06986 = 3.
We call log of base e, a natural logarithm because every other logarithm table represents a special case of this more general assembly of numbers and every other catalogue of log tables involves a complication factor of n ¹ 1, which this one single set can avoid and as we have seen even the factor n¹ 1 can still be understood in terms of the number, e.
We can and should run around and make up our precious base ten log tables because
that’s useful to our human mindset. There’s nothing so simple about the idea of
counting in multiples of 2.718281828...
Until, that is, you come to examine the problem of the unknown power and then oddly enough you have found a number which is as important to algebra and calculus and probability theory as the number p is to circles, spheres, curves and angles.
The number e is a very special number in mathematics, deep in the subject and not something humans could have invented for themselves any more than humans could have fixed the ratio p . Partly the number e is admired because it and it alone makes true the statement that n = 1.00000... That is not to say that it is an easy number to calculate with by hand, it is surely not.
So we can now guess at a very interesting law of mathematics and write this about any base, b, which is not e, we have already found rule A)
A) Log b [1+ kn ] = k. (for small k only)
Where we have now also noticed that we can express our bolded corrective factor n in terms of the natural base, e, as well. This is now rule B),
B)
n = Log e b
We get by substitution of B) into A), rule C)
C) Log b [1+ k (Log e b)] = k. (for small k only)
Which we can translate as,
bk = 1 + k (ln b) (small k only)
Which tells us, in effect, that when our powers are very small, it makes no difference what our base b is, it can be anything, our correctional term of n will always depend on two numbers, b and the mathematical constant , e = 2.718281828... and the corrective term n, now becomes a power n which, by sheer weirdness, also makes true en = b. That is to say that even the corrective factor, n, when it is not equal to 1, still has an association with the number e. So you cannot escape this number e in using logarithms. Even when you try to use other bases, it still appears in the corrections!
The above law is also an incorrect law in fact, which becomes more and more correct the smaller we make k. As k approaches zero, the above law gets more and more accurate.
In fact there is a mathematical technique for writing this kind of relationship, where numbers ‘approach’ other numbers. This branch of mathematics called limits, was developed for calculus, where this slippery idea of numbers approaching other numbers is used all of the time.
So there is indeed a method to ‘perfect’ this above expression, there is a way to make it a proper and true law; the method of using ‘limits’. It isn’t worth learning all about limits right now just to understand logarithms, however. Only if you go on and learn calculus do you need to worry about such things as limits. For now I note for the sake of sheer completeness in the discussion, that the expression,
Lim {Log b [1+ k (Log e b)] } = k.
(k → 0)
or the translation,
Lim (bk) = 1 + k (ln b)
(k → 0)
Is a fully correct law of logarithms which holds completely true and is not in any sense wrong, nor in need of any further correcting. Okay, so that’s where the base of e comes from. Now that we have seen why such a long and complicated number is called a natural base, we should talk about alternative notations because they are entirely common.
4.2.1 Alternative
Notations – log e x and ln x
Now rather than write, log e x = p, (which means ep = x) all the time, they simplified logs to the base of e, they revered log e so much that they gave this basic and naturally unnatural set of log tables, a special and distinctive, simplified set of notations, so we can write,
ln x
= loge x = p
So that the expression of three symbols, ln x, actually means the same thing as a natural logarithm which you can also think of as a natural power, where the base is in the natural ‘unit’ of e = 2.718281828... and the resulting number is some power called p and we don’t know the value of this natural power, p, but we’re able to always find it out by knowing x and knowing the location of the ln x button on our calculator. Below, of course, x = 5107.
Problem. Solve 5107 = 2.718281828... p
If I set myself this problem then I have also said,
Problem. Evaluate p = ln
5107
Which is, what power, p, must I raise e = 2.718281828... to, in order that I make the number 5107? How many times must I multiply e by itself to get 5107? Any rough guesses? About 12 times is my first guess. We type the number 5107 and hit our ln button... even my desktop Microsoft Windows calculator does this on scientific mode...
P = 8.5383674... so I was quite a way off there because I used 2 as a rough approximation to e which was very mistaken, I know that 212 is the power of 2 closest to this number 5107 (212 = 4096). To find out how bad my estimate was, let’s find out what e12 is.
e12 = 162754.791...
So I was off by a factor of 31.8, which is not so good. What is good is that I can always calculate that which I cannot estimate! In order that my estimates are exposed for the horrors that they are, I am forced to do things more exactly. It’s no crime to be so wrong in estimating.
What about a graph of ex = y and comparing that to the function ln x = y?
An ex button is also found on most standard calculators as the inverse of ln x. So you can locate values for ex and ln x without any real trouble.
4.2.2 Alternative Notations – ea
and exp(a)
Now there is another way that mathematicians came to write a number to the base of e. They found that in physics, this base of e is so useful that it winds up being used in a lot of different laws. Mathematicians too wanted to be able to raise e to complicated powers.
But how do you write this idea. You can write e(a+b+c) = n. But here the power involves a three part sum. More complicated possibilities get ugly very quickly. So what they did was invent another notation for raising e to some power.
e (a+b+c) = exp(a+b+c)
This looks complicated at first until you realise that wherever you see the term exp(a) you are really only saying raise the number e, to the power a. If you had exp(b), you would instead write exp(b) = eb.
So if I have some very long and complicated power of e, like,
exp(1+ x5 +√(3+x2+x3) + 4) = H
We have now written e raised to some horrific looking power, where rather than raise the power to the base e, we write that same idea as exp(b). It means that when we write equations for books and so on, we needn’t be so bothered with tricky formatting problems.
So exp(a) always means ea, no matter what a is ultimately equal to. It’s the same thing, just another notational method. This is nice to know because it means that when you see exp(a) in some person’s chemistry thesis, you already know what it means as a mathematical proposition. It means they wanted to write a function of e, using a power, but they found that the power expression for (a) was probably a little bit too complicated to write in the form of e(a).
When we write exp(a), the term a is a power and the base is e. That’s all there is to it.
Problem Set A) all results are rounded to 6d.p.
Problem 1) 8 = 10L L
= Log 8 =
0.903090...
Problem 2) Log10 2649 =
Log 2649 = 3.423082...
Problem 3) 928 = 10x x =
Log 928 = 2.967548...
Problem 4) Log10 359 = Log
359 = 2.555094...
Problem 5) 3589 = 10p p
= Log 3589 = 3.554973...
Problem 6) Log10 2588 =
Log 2588 = 3.412964...
Problem 7) 4563 = 10g g = Log 4563 = 3.659250...
Problem 8) Log10 25 =
Log 25 = 1.397940...
Problem 9) 2123 = 10L L
= Log 2123 = 3.326950...
Problem 10) Log10 659875 =
Log 659875 = 5.819462...
Problem Set B)all results are rounded off to nine d.p.
Problem 1) Solve for n, 7n = 6009
Log 10 7 = 0.84509804
Log 10 6009 = 3.778802204
Hence, 10(0.84509804)n = 10(3.778802204)
And n = 3.778802204 ÷ 0.84509804 = 4.471436478
Problem 2) Evaluate, Log 2 28
Log 10 2 = 0.301029996
Log 10 28 = 1.447158031
Hence, 10(0.301029996)n = 10(1.447158031)
And n = 1.447158031 ÷ 0.301029996 = 4.807359422
Problem 3) Solve for c, 1970 = 33c
Log 10 33 = 1.51851394
Log 10 1970 = 3.294466226
Hence, 10(1.51851394)c = 10(3.294466226)
And c = 3.294466226 ÷ 1.51851394 = 2.169533081
Problem 4) Evaluate, Log 39 1964
Log 10 39 = 1.591064607
Log 10 1964 = 3.293141483
Hence, 10(1.591064607) n = 10(3.293141483)
And n = 3.293141483 ÷ 1.591064607 = 2.06977232
Problem 5) Solve for f, 1943 = 28f
Log 10 28 = 1.447158031
Log 10 1943 = 3.288472801
Hence, 10(1.447158031) f = 10(3.288472801)
And f = 3.288472801 ÷ 1.447158031 = 2.272366065
Problem 6) Evaluate, Log 18 1985
Log 10 18 = 1.255272505
Log 10 1985 = 3.297760511
Hence, 10(1.255272505) n = 10(3.297760511)
And n = 3.297760511 ÷ 1.255272505 = 2.627127176
Problem 7) Solve for g, 243 = 9g
Log 10 9 = 0.954242509
Log 10 243 = 2.385606274
Hence, 10(0.954242509) g = 10(2.385606274)
And g = 2.385606274 ÷ 0.954242509 = 2.5
Problem 8) Evaluate, Log11 125
Log 10 11 = 1.041392685
Log 10 125 = 2.096910013
Hence, 10(1.041392685) n = 10(2.096910013)
And n = 2.096910013 ÷ 1.041392685 = 2.013563224
Problem 9) Solve for φ,
2127 = 15φ
Log 10 15 = 1.176091256
Log 10 2127 = 3.32776749
Hence, 10(1.176091256) φ = 10(3.32776749)
And φ = 3.32776749 ÷
1.176091256 = 2.829514686
Problem 10) Evaluate, Log13 256
Log 10 13 = 1.113943352
Log 10 256 = 2.408239965
Hence, 10(1.113943352) n = 10(2.408239965)
And n = 2.408239965 ÷ 1.113943352 = 2.161905235
Problem Set C)
Problem 1) 7n = 6009 then
this becomes n = Log 7
6009
Problem 2) if b = Log 2 28 then
this becomes 2b = 28
Problem 3) 1970 = 33c then
this becomes c = Log 33
1970
Problem 4) if b = Log 39 1964 then this becomes 39b =1964
Problem 5) 1943 = 28f then
this becomes f = Log 28
1943
Problem 6) if b = Log 18 1985 then this becomes 18b = 1985
Problem 7) 243 = 9g then
this becomes g = Log 9
243
Problem 8) if b = Log11 125 then
this becomes 11b =
125
Problem 9) 2127 = 15φ then
this becomes φ = Log 15 2127
Problem 10) if b = Log13 256 then
this becomes 13b =
256