# Moving on to Higher Powers - a First Look at Exponents

## The Power Of Many

Having finished with those concepts we found in And then there were Three - a Study on Cubes, our best bet now is to look at** the Laws of Exponents**, and find out what we can and cannot get away with when dealing with bases raised to certain powers.

We see that for example

**3 ^{5}**=

**3**×

**3**×

**3**×

**3**×

**3**,

which can also be seen as **3² **×** 3³**, which also equals **3 ^{( 2 }**

^{+}

**. Then there are ones like (**

^{ 3 )}**3³**)

**²**or (

**3²**)

**³**, both of which are

**3**

^{(2}^{×}

**, or**

^{3)}**3**. With the first one you can expand it to read

^{6}(** 3 **×** 3 **×** 3** )**²**, or even

( **3** × **3** × **3** ) × ( **3** × **3** ×**3** ),

while the second one would be rendered as :

(** 3 **×** 3** )**³** = ( **3** × **3** )** **× ( **3** × **3** )** **×( **3** × **3** )** **

Both of these, when you remove the parentheses, end up as what we stated in the first place ; **3 ^{6}**.

Now there seems to be some confusion as to why any number ( other than **zero **), raised to the power of **zero**, should equal **one**, among other such things. But never fear - all these facts *are* consistent with our** Laws of Exponents**, which are given below.

## Exponents in Action

## Looking at How the Powers Behave

It was these simple rules which led **John Napier** (**1550 - 1617**) to the invention (or discovery) of **Logarithms** in the early Seventeenth Century, and we shall also be having a look at that soon enough. But first, think about how these hard and fast rules, which always work, give us the ideas we now have.

Consider the statement that

**a ^{x}**÷

**a**=

^{y}**a**

^{(x }^{-}

**.**

^{ y)}Now if **x ** and **y** have the same value, such that **y **=** x**, then, as an example, we would have something like : ** 3² **÷** 3² **=** 9** ÷** 9 **=** 1**, and, according to the** Laws of Exponents,** this would be **3 ^{(2 – 2)}**, which of course would equal

**3º**, so that

**3º**does therefore equal

**one**, simply because it works out that way.

Then there is the idea that **a ^{(- x)}** =

**1**÷

**a**, so that, if we went something like :

^{x}**3**, this, according to our general rule, is supposed to equal

^{( - 1 )}**1**÷

**3¹**, which just means

**⅓**. To prove this, let us consider the expression

**3¹**÷

**3²**. According to one of the other rules, this is the equivalent of saying

**3**, which also equals

^{(1 - 2)}**3**. Now

^{(- 1)}**3¹**÷

**3²**can also be expressed as

**3**÷

**9**, which itself equals

**⅓**, and thus this shows that

**3**

^{(-1)}*is*the equivalent to saying

**⅓**, which then proves our point, that

**ba**does indeed equal

^{(- y) }**b**÷

**a**.

^{y}Next, if we also look at how **a ^{(x÷y)}** equals the

**y**root of

^{th}**a**, it also means that

^{x}**a**. Well, this is shown to be true if we consider, by virtue of another rule, how that if

^{½}= √( a )**( a**,

^{½})² = a^{( ½ × 2 )}**= a¹**, which equals

**a**, then

**a**

^{½}=

**√(a)**, since if by squaring

**a**you get

^{½}**a**, this means that

**a**, must also be equal

^{½}**to**

**√(a)**.

There are in addition to the ones above, a good number of other rules which need to be considered :

If **a ^{– (x÷y) }**=

**b**, then

**b**, and (

^{– (x÷ y)}**1**÷

**b**)

**=**

^{(y÷x)}**a**, and by the same token if the above fractional exponents were positive, the same rules would apply.

Also : If **a ^{x }**=

**b**, then

**b**=

^{(1÷x) }**a**, and (

**1**÷

**a**)

**=**

^{– b }**a**

^{b}And one last one we can consider here is the fact that (** a** -** 1 **)**a ^{n }**

**+ a**. For example, we see the following :

^{n}= a^{(n+1)}**2**×

**3²**+

**3²**=

**3³**.

Now we know that in algebra we can *multiply* the same base raised to the same or different powers by having the one base raised to the sum of the two powers. For example :

**5²** × **5³** = **5 ^{( 2 + 3 ) }**=

**5**

^{5 }

But what if we wanted to *add* two such bases raised to powers ? Well, interestingly enough, as much as you need to add powers in order to multiply bases, you may also multiply in order to add bases.

The formula for all of this stands as follows :

**x ^{a }**+

**x**

^{y}= (

**x**+

^{( y - a )}**1**)

**x**

^{a}or (

**x**+

^{( a - y)}**1**)

**x**

^{y}This works whether **a** or** y** is greater, and even if it is *positive* or *negative*.

For example : **5²** + **5³** = (** 5 ^{( 3 – 2) }**+

**1**) ×

**5²**

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

**1**) ×

**25**=

**6**×

**25**

= **150**, or ( **25** + **125** )

Now it doesn’t matter which of the two exponents is symbolised by which of the two superscript letters, **a** or **y**, since we still get the same answer even if they are reversed :

**5²** + **5³** = ( **5 ^{( 2 – 3 ) }**+

**1**) ×

**5³**

= (** 5** ^{- 1} + **1** ) × **125** = **1.2** × **125**,

= **150** = ( **25** + **125** )

And this works as long as, if we change the value of one letter, we do so to the other.

Now we see how all this works by simple algebra when we expand the brackets :

(** x ^{( y - a )}**+

**1**)

**x**=

^{a}**x**×

^{a}**x**+

^{( y - a )}**x**×

^{a}**1**=

**x**+

^{( a + y - a )}**x**

^{a}Which gives us **x ^{y}**+

**x**

^{a}, same for both equations.

The addition formula can even be rewritten a third way, and this is that we can say :

## More on Exponent Behaviour

This could be a simpler way to work the question out, as long as you can work out that the value of one of these superscript letters represents the *difference* between one exponent and the other, and only becomes an exponent itself in the formula.

Since we saw in our first example, that the two exponents were in fact **3** and **7**, all we did was turn **7** into ( **3** + **4** ), because **4** is the *difference * between **3** and **7**.

For example : **3²** + **3³** also can be written **3²** + **3 ^{( 2 + 1)}**

Let’s say **a **=** 2** and **y **=** 1** ( not** 3**, but **1**, since **1** is the *difference* between **2** and** 3** )

Then we have :

( **x ^{1}** +

**1**) ×

**x**

^{2 }=

**4**×

**9**=

**36**, which is

**9**+

**27**

Now all of this is because the difference between a number raised to a certain power, and the *same* number raised to the *next* power, is the number raised to its *first* power, *plus** * one less than the number, times itself raised to this * first* power.

For example : **3³** = **3²** × **3**, or **3²** + ( **3** - **1** ) × **3²**

**7 ^{4}** =

**7³**×

**7**, or

**7³**+ (

**7**-

**1**) ×

**7³**

Now this leads to an interesting discovery, for if you wish to increase the difference *between* exponents, so that you can find the * difference* between the values of the numbers raised to them, you then need to

*increase*what you multiply the term within the parentheses by :

**3 ^{4}**=

**3²**×

**3²**or

**3²**+ (

**3²**-

**1**) ×

**3²**

**3 ^{5}** =

**3³**×

**3²**or

**3²**+ (

**3³**-

**1**) ×

**3²**

**3 ^{5}**=

**3²**×

**3³**or

**3³**+ (

**3²**-

**1**) ×

**3³**

So, generally speaking, we have :

**x**^{( a + y )} = **x ^{a}**×

**x**=

^{y}**x**+ (

^{a}**x**-

^{y}**1**) ×

**x**

^{a, }or

**x**+ (

^{y}**x**-

^{a}**1**) ×

**x**

^{y}All very well and good, but then what if we were wishing to *subtract*, rather than add ?

**x ^{a }**-

**x**= (

^{y}**x**-

^{ (a - y)}**1**)

**x**

^{y}**x ^{y}**-

**x**= (

^{a }**x**-

^{ ( y - a )}**1**)

**x**

^{a}This time there is only one equation for each thing we wish to do, and it is to be noted, that

most of the time, **x ^{a }**-

**x**will

^{y}*not*be the same as

**x**

^{y }-

**x**

^{a }, so these are not always interchangeable.

Let’s see some more examples :

**5³** - **5²** = (** 5 ^{( 3 - 2 )}** -

**1**) ×

**5²**=

**4**×

**25**=

**100**, while

**5²** - **5³** = ( **5 ^{( 2 – 3 )}**-

**1**) ×

**5³**=

**- .8**×

**125**=

**- 100,**both of which are in fact true.

## Comment

It certainly is interesting that it works out this way, and this is not the only discovery I have made. Looking at it from another viewpoint, seeing we can get two powers of the same base to add by multiplying, what then can we do to *different* bases of the *same* power ?

If we add two *cubes*, we can in fact evenly divide this answer by the sum of the two numbers we have cubed.

And as a matter of fact, this *sometimes* works for any number of bases raised to the same power, then added together, like three, four, five, etc., and we shall study that in depth a bit later.

Now so much for *cubes*, what about __squares__, then ? Let’s check these out as well :

**1²** + **2²** = **1** + **4** = **5**, which is *not* divisible by the numbers (** 1** + **2** ), and as a matter of fact, only two numbers raised to *odd* powers give an answer evenly divisible by the sum of the two base numbers.

This means only powers of one, three, five, seven, nine, and so on will give an answer evenly divisible by the sum of both bases. Why then does this little old thing work that way ? *What* determines the number that we can multiply the sum of both bases by, to get our sum of both bases raised to our *odd* numbered power ?

## That is to say, knowing that

Where do we get the sevens from for examples one and two, and the three for example three ?

Well, it is all because of the *difference* between **x³** + **y³**, and ( **x** +** y** )**³**. Now *don’t * go thinking that **x³** + **y³** and ( **x** + **y** )**³** are the same - they’ll throw chalk at your ears for trying to pull that one.

Look at ( **x** + **y** )**³**, as it is, when expanded :

( **x** + **y** )**³** = **x³** +** 3x²y** + **3xy²** + **y³**

Well, will you look at that ! ( **x** + **y** )**³** is actually equal to **x³** + **y³** *plus* **3x²y** + **3xy²**, so they certainly are *not* the same. Now, an *addition* is not much good to us, since we wish to find the relationship between **x³** + **y³** and **x** +** y**, which is a *factor * of it. The thing is, as much as **x** +** y** is a factor of **x³** + **y³**, it is also a factor of ( **x** + **y** )**³**, and this gives us a potential reason as to why** x** + **y** is also a factor of **x³** + **y³**, so let’s have a look.

## An Answer to the previous Question

( **x** + **y** )**³ **= (** x** + **y** )(** x** + **y** )(** x** + **y** )

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

Now, as we stated before, we see that :

( **x** + **y** )**³ **= **x³** +** 3x²y** + **3xy²** + **y³**,

and if we look carefully at this, we see :

**x³** + **y³** = ( **x** + **y** )**³ ** - ( **3x²y** + **3xy²** )

= ( **x** + **y** )**³ ** - **3xy**( **x** + **y** ),

Since if you expand the above second term out, you do get the one above it.

**x³** + **y³** therefore = **( x + y )³** - **3xy**( **x** + **y** ), and you notice both terms here contain (**x** + **y**), so we can now make it one factor of an equivalent expression thus :

**x³** + **y³** therefore = **( x + y )³** - **3xy**( **x** + **y** ), so

**x³** + **y³** = ( **x** +** y** ) [ ( **x** + **y** )**²** - **3xy** ], thus proving that indeed** x³** + **y³** *is* divisible by (**x** +** y**), meaning we can also work out what the factor of (**x** +** y**) will be, when we put our numbers into its formula :

[ ( **x** + **y** )**²** - **3xy** ]

## Why it is So

Well, in our first case, when we went and had

**1³** + **3³** = **1** + **27** = **28** = ( **1** + **3** ) × **7**, we know then that [ ( **x** + **y** )**²** - **3xy** ] = **7**. Actually, looking at our formula, it doesn’t look like it matters which number is **x**, and which one shall be **y**, since within the formula, both are added together, then squared, and it matters not in which order we add, and in the second part, both are multiplied together by three, so which ever order we use does not matter. For argument’s sake, let’s say **x** = **1**, and **y** = **3**.

Now if this is so, shall we put these values into the formula, then ?

[ ( **x** + **y** )**²** - **3xy** ]

[ ( **1** + **3** )**²** - **3** × **1** ×** 3** ] = **16** - **9** = **7**,

Thus, for this one at least, it works. Remember then that here we are finding the number that multiplies by the sum of our two bases to give us the sum of the *cubes* of the same two bases.

Let’s try the second one :

**2³** + **3³** =** 8** + **27** = **35** = ( **2** + **3** ) ×** 7**

**x** = **2**, **y** = **3**

[ ( **x** + **y** )**²** - **3xy** ]

[ ( **2** + **3** )**²** - **3** × **2** × **3** ] = **25** - **18** = **7**, and once again, we rock ! Well, third time’s a genius :

**1³** + **2³** = **1** + **8** = **9** = ( **1** + **2** ) × **3**

**x** = **1**, **y** = **2**

[ ( **x** + **y** )**²** - **3xy** ]

[ ( **1** +** 2** )**²** - **3** × **1** × **2** ] = **9** - **6** = **3**

And now we know why and how the sum of two *cubes* can be evenly divided by the sum of their bases, using a clever little bit of algebra.

## On to the very Point

Now the last expression from before can become

**x³** - **y³** = ( **x** - **y** )**³** + **3xy**( **x** - **y** )

( Because again if we expanded the polynomial just above, we would end up with the expression we had before. ) **x³ **- ** y³ **= ( **x **-** y **)**³** +** 3xy**( ** x **- **y **)

So that again, like before, we see that both terms in this equation contain ( **x **-** y **), and we can therefore rearrange the whole thing so that as much as **x³ **-** y³ **=** **(** x **-** y **)**³** +** 3xy**(** x **-** y **), we see that it also ** **= ( **x** - **y **) [ ( **x **- **y **)² + **3xy** ] So, for example :

5³

-** 2³ **=** 125 **-** 8 ** = **117 **= ( **5 **-** 2 **) × **39**, so that the formula for the **37** we multiply ( **5 **-** 2 **) by is this one : [ ( **x **-** y **)² +** 3xy** ]

Now this time because we are subtracting, it *does* matter which order we have the numbers in, so because **2 **is being subtracted from **5**, then **x **= **5**, and so **y **=** 2**. Then we substitute into the formula :

[ ( **x **-** y **)² +** 3xy** ]

[ (** 5** -** 2 **)**² **+** 3 **×** 5 **×** 2 **] =** 9 **+** 30 ** =** 39**

Faith and begorrah ! For sure and it works !** **Well, there is plenty more to come, as we continue this in the next Hub.

## Final Thought

We continue this journey in the next Hub, The Power of Many More - more on the Use of Exponents, and 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, Mathematics - the Science of Patterns , 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 :

## Disclaimer

Although some of the content of this Hub is Mathematical knowledge found in the public domain and therefore of common knowledge and not subject to Copyright, other information will have 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. Some can also be found through sites like Wikipedia ( Copyright 2013, the Wikimedia Foundation ), which is a good source of information. Some 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.