Moving on to Higher Powers - a First Look at Exponents

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

3⁵= 3 × 3 × 3 × 3 × 3,

which can also be seen as 3² × 3³, which also equals 3^{( 2 + 3 )}. Then there are ones like (3³)² or (3²)³, both of which are 3^(2×3), or 3⁶. With the first one you can expand it to read

( 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 ; 3⁶.

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.

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^x, so that, if we went something like : 3^{( - 1 )}, this, according to our general rule, is supposed to equal 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^{(1 - 2)}, which also equals 3^{(- 1)}. Now 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^{(- y)} does indeed equal b ÷ a^y.

Next, if we also look at how a^(x÷y) equals the y^th root of a^x, it also means that a^½ = √( a ). Well, this is shown to be true if we consider, by virtue of another rule, how that if ( a^½ )² = a^{( ½ × 2 )}, = a¹, which equals a, then a^½ = √(a), since if by squaring a^½ you get a, this means that a^½, must also be equal 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 ^{– (x÷ y)}, and ( 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 )aⁿ + aⁿ = a⁽ⁿ⁺¹⁾. For example, we see the following : 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 )} = 5⁵

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

= ( 5¹ + 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 :

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 :

( x¹ + 1 ) × x² = 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

7⁴ = 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 :

3⁴= 3² × 3² or 3² + ( 3² - 1 ) × 3²

3⁵ = 3³ × 3² or 3² + ( 3³ - 1 ) × 3²

3⁵= 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^ywill 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.

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 ?

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.

( 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 ]

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.

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.

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 :

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