# The Power of Many More - more on the Use of Exponents

## More on the Issue of Dividing sums of Powers

We carry on from the previous Hub, Moving on to Higher Powers - a First look at Exponents

, where we looked at the divisibility of sums of numbers raised to a given power, and move on now to explore this idea a bit further. But first, you may have remembered me saying that the process shown in the previous Hub only works for *odd* numbered powers, and in my example of adding two __squares__ together, we ended up with the following :

**1²** +** 2² **=** 1 **+** 4 **= ** 5**, which is ** not** divisible by the numbers (

**1**+

**2**). Now this is quite clear, but one example is not enough to prove a point, so let’s see why it is so in more general terms.

Since we showed that the sum of two *cubes* was divisible by the sum of their bases when we spoke of the difference between **x³ **+** y³ **and (** x **+ **y **)**³**, we can actually try the same sort of monkey business with the difference between **x² **+** y² **and ( **x **+** y **)**² **

( **x **+** y **)**² **=** x² **+** 2xy **+** y²**, we can see then that **x² **+** y² **= ( **x **+** y **)**² **-** 2xy**

Now for us to get a whole number answer if we were to divide **x² **+** y² **by** x **+** y**, then **2xy** would have to in some way be divisible by (** x **+** y **), but it is not. Its factors are ** 2**, **x**, and **y**, not a penny more, not a penny less. That’s all, folks. So this means that the sum of two __squares__ will end up being the __square__ of the sum of their bases, minus twice their product :

**x² **+** y² **= ( **x **+** y **)**² **-** 2xy**

Of course we see that the first term of this expression, ( **x **+** y **)**²**, is evenly divisible by (** x **+** y **), since ( **x **+** y **)**² **=** ** ( **x **+** y **)(** x **+** y **), but the other part, being the **2xy** subtracted therefrom, is not, at least not necessarily, and the only way it could be would be by some coincidence, whereas, for the most part, considering the values the variables can take, it will not be. So if we were to divide the whole lot by ( **x **+** y **), we would end up with :

[ ( **x **+** y **)**² **- **2xy **] ÷ ( **x **+** y **)

= **x² **+** y² **÷** 2xy**

= (** x **+** y **) - [** 2xy **÷(** x **+** y **) ]

## Checking it Out

So, let’s see if this is what we get with our example expression from before.

**1² **+** 2² **=** 1 **+** 4 **=** 5**, and what we want to know is, what is **5 **divided by (** 1 **+** 2 **)** **? We already know we will not get a whole number, but rather some kind of fraction

Well here again, because we add, it does not matter which number we represent by which letter, so let’s just say then that ** x **=** 1**, and **y **=** 2**.

Then, for the equation (**x **+ **y** ) - [ **2xy **/ ( **x** + **y** ) ** **]

(** 1 **+** 2 **) - [ (** 2 **×** 1 **×** 2 **) ÷ (** 1 **+** 2 **) ] =** 3 **-** ** (** 4 **÷** 3 **) =** 1⅔**,

which really does equal our answer of **5 **÷** **(** 1 **+** 2 **), since this equals **5 **÷ **3**, which then does equal** 1⅔**.

All of this was really just to show that the sum of two __squares__, or any other *even* numbered power does *not* divide evenly by the sum of its two bases, while *cubes* and all other odd numbered powers do.

**1 ^{5}** +

**3**=

^{5}**1**+

**243**=

**244**= (

**1**+

**3**) ×

**61**

The reason for all of this is something to do with how the binomials (** x** +

**)**

**y****expand :**

^{n}## By Example

Let’s see then if this works, and what we are looking for here, to be clear, is that as according to our example, the above should be the formula to obtain the number **244 **:

** 1**^{5 }+** 3 ^{5 }**=

**1**+

**243**=

**244**= (

**1**+

**3**) ×

**61**

Let ** x **=** 1** and ** y **=** 3**

= ( **x **+** y **){ ( **x **+** y **)**4 **- { **5xy**[ ( **x **+** y **)**²** - **3xy **] } + **10x²y² **}** **so that we have :

**1 ^{5 }**+

**3**= (

^{5}**1**+

**3**) × {(

**1**+

**3**)

**4**-

**{**

**5**×

**1**×

**3**× [ (

**1**+

**3**)

**²**-

**3**×

**1**×

**3**]} +

**10**×

**1²**×

**3²**},

which equals

**1 **+** 243 **=** 4 **×** **{** 44 **- [ **15 **×** **(** 16 **-** 9 **) + **90 **]}

**244 **= **4 **×** **[** 256 **-** **(** 15 **×** 7 **+** 90 **) ] =** 4 **×** ** ( **256 **- ** 195 **) = **4 **× ** 61**, which is correct.

## Explanations

If we leave the (** x **+** y **)** **part as a factor in the formula, we will end up with the number we need to multiply our (** x **+** y **) by in order to find **x ^{5 }**+

**y**. The important part is to have all the brackets the right way round, and carry out our operations in the correct order. Let’s have one more go to check our formula.

^{5}**2 ^{5 }**+

**5**=

^{5}**32**+

**3125**=

**3157**= (

**2**+

**5**) ×

**451**

Another significant number, for you should know this is the temperature in **degrees Fahrenheit**, at which paper burns.

**2 ^{5 }**+

**5**=

^{5 }**32**+

**3125**=

**3157**= (

**2**+

**5**) ×

**451**

= ( **x **+** y **){ ( **x **+** y **)** ^{4}**- {

**5xy**[ (

**x**+

**y**)

**²**-

**3xy**] } +

**10x²y²**}

Again, no never mind as to which number is which, so let’s say **x **=** 2** and **y **=** 5**

(** 2 **+** 5 **) × {( **2 **+** 5 **)** ^{4}**- {

**5**×

**2**×

**5**× [ (

**2**+

**5**)

**²**-

**3**×

**2**×

**5**]} +

**10**×

**2²**×

**5²**}

= **7 **× {** 7 ^{4 }**- {

**50**× [

**7²**-

**30**] +

**1000**}

= ** 7 **×** **{** 2401** - [ **50** × **19 **+** 1000 **] }

=** 7** × {** 2401 **-** 1950 **} = **7 **×** 451 **=** 3157**

So it certainly does all work. Now believe me, the rest of the expansions of **binomials **raised to greater powers will be even more complicated, but whatever the case, there is some kind of formula that can be worked out to predict the value of either the sum of two numbers raised to the same power, or the result you get when you divide such a sum by the sum of the two bases involved, which is to say, (** x **+** y**)^{n }÷** **(** x** +** y **), and we need know is that sometimes it divides evenly, sometimes not.

## Practicalities of the Exponents

Now, as mentioned before, we can also add together *more than* **two **bases raised to an add numbered power, to get an answer divisible by the sum of those **three **or more bases.

**1³** +** 2³ **+** 3³ **=** 1 **+** 8 **+** 27 **=** 36 **= (** 1 **+** 2 **+** 3 **) ×** 6**

**2³** +** 3³ **+** 4³ **=** 8 **+** 27 **+** 64 **=** 99 **= (** 2** +** 3 **+** 4 **) ×** 11**

But this is only as long as the bases you are adding are in some patterned order. For example, in the ones we used above, both sets of numbers were consecutive, and any number of consecutive bases will work :

Also, if each base is the same number :

So if there are an identical number of each base, you will be right. This is why the polynomial expansion of ( **x **+** y **+** a **)**³ **doesn’t allow us to divide any part of it by ( **x **+** y **+** a **)** **in the same way we did for the binomial.

That is, we can say that **x³ **+** y³ **+** a³ **=** **(** x **+** y **+** a **)**³ **- ( **3a²y **+ **3x²y **+** 3y²a **+** 3y²x **+** 3a²x **+** 3x²a **+** 6axy **), and while we can get ( **x **+** y **+** a **)**³ **to divide evenly by ( **x **+** y **+** a **), we cannot get the second part to do so, since no matter how we manipulate it, we cannot get it to equal one polynomial as multiplied by ( **x **+** y **+** a **).

Obviously, it can evenly divide into what we know as ( **x **+** y **+** a **)** **depending on which numbers we use, whether they are consecutive or multiples of the same number, or in some other sort of pattern, but in those cases what we end up as the factor of ( **x **+** y **+** a **)** **that multiplies to give us **x³ **+** y³ **+** a³ **can be expressed in some other different sort of way.

There is another way we can get our expansion to give us factors that evenly divide, and that is not to express it as a trinomial, but as an expression involving **one **unknown, where what would be the other two unknowns are expressed in terms of the first.

## Another Example

Let’s take the example of each unknown being a multiple of **three**, and express it a different way :

**3³** +** 6³ **+** 9³ **;** x ** = **3**, so this is **x³ **+** **(** 2x **)**³ **+ ( **3x **)**³**

and this is *supposed to be* divisible by (**x **+** 2x **+** 3x **)**³**

Thus we have with all within parentheses worked out to be equal to **x³ **+** 8x³ **+** 27x³**,

which divides into (** 6x **)**³ **=** 216x³**

Now we can actually see by the values we have given multiples of **x**, that this will work, since **x³ **+** 8x³ **+** 27x³** =** 36x³**, which is **one sixth **of **216x³**, but we need to find a general expression that can be used for any multiples of **x**.

Since the gap between all three numbers is the same, we can express them in terms of the common difference between them, by having :

## Put another Log on the Fire

We look now at **Logarithms** - a valuable tool in **Post Medieval England**, without electronic calculators, and such forth. Let’s have a wee look at them, and see how this is so.

If **y = a ^{x}**, then

**log**. This tells us that what we call the

_{a}y = x**Logarithm**is basically an

**exponent**, a

**power**, to which a

**base**is to be raised to give us an answer. Our books of tables and scientific calculators give us only

**two bases**to work in, the

**Common Logarithm**to the

**Base Ten**, and the

**Natural Logarithm**to the

**Base**of

**e**. We shall see more of the latter at a later date, but our concern for the moment is those

**Logs**we can use that are generated from the

**base**of

**ten**.

Now the basic principle behind **Logarithms** is the idea, that to multiply numbers of the same base, you add their exponents : **a ^{x} × a^{y} = a^{(x + y)}**, so that, as long as you

**dealing with the same**

*are***base**only, you’re sweet. It’s all good. This then is why the

**Log**tables are all those of the one

**base**each time.

The basic **Log **rules then, are as follows :

**log x** + **log y** = **log** (**xy**) example : **log 2 + log 3 = log 6 **

**log x - log y = log (x ÷ y) **example : **log 8 - log 2 = log 4 **

**log x ^{n} = n log x** example :

**log 2³ = 3 log 2**

Now, because of this last one, we actually have what I like to call the **Change of Base Formula :**

**log _{b} x** = (

**log**÷

_{a}x**log**), and this is useful seeing we are normally using

_{a}b**ten**as a

**base**. What we do, is, if for instance we wanted to know the

**logarithm**to the

**base**of

**4**of the number

**20**, which is to say, to which power do we raise

**4**in order to get

**20**?

In this particular case, **b = 4**, **x = 20**, and **a = 10**, so all we do is divide the **Common Logarithm **of **20** by the **Common Logarithm** of **4**. Incidentally, the answer we get is **2.161 ( rounded up to 3 d.p. )**, and what this basically means, is that **42.161 = 20**.Now, as to how we arrived at such a formula, we used one of the other **log **rules to help work it out. This is interesting, that both the **power** and **log** rules all work together to prove each other.

Now we are saying that **logb x = ( loga x ÷ loga b )**, but why ? Let’s look at an example :

**log _{5} 125 **=

**3**( which is a fancy way of saying

**5³ = 125**)

So, according to this rule we are looking at :

**log _{5} 125 **= (

**log**÷

_{a}125**log**)

_{a}5Now, by another rule, if **log 5³ = 3 log 5**, then :

**log 5 ^{⅓}** =

**⅓ log 5**= (

**log 5**) ÷

**3**

therefore, **log 125 ^{⅓}**, ( which also equals

**log 5**, since

**5**is the

**third root**of

**125**) = (

**log 125**) ÷

**3**.

So, since **log 5 **= (** log 125 **) ÷** 3**, then we can say that **( log 125** ÷** log 5** ) =** 3**, by a simple rearrangement of terms.

Now **3 **=** log _{5} 125**, and since as we’ve just seen, it also = (

**log 125**÷

**log 5**), then

**log**also equals (

_{5}125**log 125 ÷ log 5**), which is what we had set out to prove, seeing that the

**logarithms**that are in the above parentheses can be of any base.

In previous times, the use of **Logarithms **was more widespread and necessary than it is now, with our flash electronic abacuses. **Logs** were used in the old slide rules to aid in multiplication. They do still, however, have a good deal of usefulness even in this modern age, so a reasonable knowledge of their workings will certainly not go astray.

## Fermat’s Last Stand

One of the most enduring mathematical mysteries – that is before it was solved about twenty years ago by the British mathematician Andrew Wiles, was known as **Fermat’s Last Theorem**. Over three hundred years ago, this seventeenth century French lawyer and mathematics genius, Pierre de Fermat ( 17 August 1601 – 12 January 1665 ), made a statement in the margin of a notebook that he had proven a famous problem. This was about the addition of two whole numbers raised to the same power, and whether their sums could also produce a value that could also be the result of a third whole number raised to that same power.

That is to say, if we can have **3² **+** 4² **=** 5²**, so that two __squares__ can be added to equal a** third **__square__, can we add two *cubes* together to find a number, the sum of these two *cubes*, which is also the *cube *of a whole number itself. So, could we say that **a³ **+** b³** =** c³**, where **a**, **b **and **c** are all whole numbers, or in fact do so with any power *higher* than **two**?

The answer is no. Although we can add two __squares__ to make another __square__, we *cannot* add two *cubes*, nor **two **of any higher power, to add up to one of the same, such that **a ^{n}** +

**b**≠

^{n}**c**if

^{n}**n**>

**2**, and

**a**,

**b**, and

**c**, are whole.

It must be noted though, that you can add **three **or more *cubes* together to get another *cube*, or *more than* two of any other power, but you just cannot add **two** of any power other than **two** itself to get a sum equal to a whole number evenly raised to that same power.

Get it ? You think so ? Great ! It’s really quite simple to understand the overall idea - although the proof of **Fermat’s Last Theorem** is itself rather complex and advanced, and at this point not suitable for the delicate ears of little children. ( In fact, I don’t think it’s even suitable for my delicate brain, as yet. )

We will come soon enough to the story of the basic **Pythagorean Triplet**, which is a right angled triangle** **of sides with the ratio of **3** :** 4** :** 5**, where we shall see that **3² **+** 4² **=** 5²**. But before looking specifically at that, let’s see what actually comes from it first. See, if we take those three numbers, and raise them to the next power, we find that **3³** +** 4³ **+** 5³** =** 6³**. Now if we multiply all three of these first bases by the same number, we will then get this :

Now, multiplying each term in our first equation by ** two**, and then *cubing* each term in turn and adding, is the equivalent of multiplying the whole thing by **eight ** ( **2³** ). What this means is that **12³ **is **8 **×** 6³ **( or **( 2 **×** 6** )**³ **), because by * doubling* each term from **3³** +** 4³ **+** 5³** =** 6³**,** **we would have :

Now why all these ** cubes** added together happen to add up to a certain other

**, I’m not as yet sure, but the relationship is really there. We have after all seen that**

*cube***and other odd number exponents will give us numbers divisible by the sums of the bases involved, and this is true of the examples I have given, but most of the time these sums of multiple**

*cubes***will not necessarily be**

*cubes***themselves.**

*cubes*There are actually very many more ways I could manipulate numbers to show certain patterns, but a lot of them are kind of obvious, and stand to reason enough to be taken for granted.

## Let's get to the Root of the Problem

I am sure that everyone is aware of the idea that **2 **+** 2 **=** 2 **×** 2** =** 2²**. This happens to be the only whole number to behave in this way, that the result is the same when you raise it to a given power, or multiply it by the same number as the power to which you raised it.

There are, I discovered, also other numbers that do this. The book I read pointed out the first one, and then I noticed the pattern from there.

Understand that in the top row,** 2** can be rewritten as : ** ^{1}√2**, meaning that

**2**is the

**first root**of two, just as

**√3**is the same as saying

**, meaning the**

^{2}√3**second**, or

**square root**of two, and also then that just as

**can also be rendered as**

^{2}√3**3**, so then

^{½}**can be expressed as**

^{1}√2**2**=

^{( 1 / 1) }**2¹**.

Now if we wish to have a general formula for this, we can say that

(** a** +** 1 **)^{( }** ^{ 1 ÷ a )}**× (

**a**+

**1**) = [ (

**a**+

**1**)

^{(}

^{ 1 ÷ a }^{) }]

^{(}

^{ a }^{+}

^{ 1 }^{)},

which basically means that any number raised to the root of one less than that number, when multiplied by itself, will be the equivalent of raising it to the power of itself. This is because we have :

## In Conclusion

So both expressions are the same. Now if you are not clear what this means, all it is saying is that there are certain numbers, that if you multiply them a certain number of times, you get the same answer as if you raised them to the power of that same number of times. Now the numbers are always one more than the root they are manipulated to. For example, we have the square root of three, ( or two root three, if you like ), and the *cube* root of four, ( which is the third or three root of four ), and so on. We also see that in order to show its proof, we had to express these roots as numbers raised to fractional exponents, so that ** ^{2}√3** is the same as saying

**3**, while

^{½}**can also be expressed as**

^{3}√4**4**, and so on.

^{⅓}We continue this journey in the next Hub, Mathematics - the Science of Patterns .

If You are curious, take a look at the other Hubs, The Maths They Never Taught Us - Part One, The Maths They Never Taught Us - Part Two , The Maths They Never Taught Us - Part Three, The Very Next Step - Squares and the Power of Two , And then there were Three - a Study on Cubes, Moving on to Higher Powers - a First look at Exponents, More on the Patterns of Maths, Mathematics of Cricket , The Shape of Things to Come , Trigonometry to begin with, Pythagorean Theorem and Triplets, Things to do with Shapes, Pyramids - How to find their Height and Volume, How to find the Area of Regular Polygons, The Wonder and Amusement of Triangles - Part One, The Wonder and Amusement of Triangles - Part Two, the Law of Missing Lengths, The Wonder and Amusement of Triangles – Part Three : the Sine Rule, and The Wonder and Amusement of Triangles - Part Four : the Cosine Rule.

Also, feel free to check out my non Maths Hubs :

Bartholomew Webb , They Came and The Great New Zealand Flag

Just take a good look at it, and note how interesting it all is, then see if you can come up with anything else along the same lines. As usual, I would appreciate any comments, feedback and suggestions which would be given due credit, or indeed have a go and publishing Your ideas Yourselves, but firstly, by all means, add Your comments - it's a free Country.

## Disclaimer

Even though some of this Hub contains certain Mathematical knowledge that can be accessed in the public domain, and therefore is not subject to any Copyright, other information has been drawn from textbooks which themselves are Copyright, but only in the sense of how they deliver the information which itself is shared and sometimes ancient Mathematical knowledge. Other information has also been found on Wikipedia ( Copyright 2013, the Wikimedia Foundation ), which is a good source of information. Part of this is also my own discovery, but may also independently have been found by others.

Any quote or part of this material which seems to belong to any other author should be treated as such, and I claim no ownership of anything I did not myself invent or discover. The Title Put Another Log on the Fire, is that of a chauvinist anthem from 1975, written by Tompall Glaser ( 1933 - 2013 ), and sung in New Zealand by Bill and Boyd.

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