Inverse Operations

I said a little about operations that are inverses to each other when I talked about square roots. Operations (or functions on numbers – I may use operations and functions interchangeably) are inverses of each other if those operations, when performed sequentially on a number, result in the original number.

For example, consider the operations of adding, then subtracting 4 to/from a number. Let’s use 5:

5 + 4 = 9, 9 – 4 = 5, the original number. More directly,

5 + 4 – 4 = 5 + 0 = 5

So no surprise, addition and subtraction are inverse operations. It doesn’t matter what the starting number is or the number you are adding or subtracting. In general, if you start with a number x, + y and – y are inverse operations:

x + yy = x

The order of these operations does not matter – I can first subtract 4 then add 4. Multiplication and division are also inverse operations. For example,

5 × 2 = 10, 10 ÷ 2 = 5 or 5 × 2 ÷ 2 = 5 × 1 = 5

Other less familiar inverse operations are squaring a number and taking a number’s square root. For example, I can square 4 then take it square root or take its square root the square the result and I will get 4 back in either direction:

\[
\begin{array}{l}
{\sqrt{{4}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{16}\hspace{0.33em}{=}\hspace{0.33em}{4}}\\
{{\left({\sqrt{4}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}^{2}\hspace{0.33em}{=}\hspace{0.33em}{4}}
\end{array}
\]

In general,

\[
\begin{array}{l}
{\sqrt{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{x}}\\
{{\left({\sqrt{x}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{x}}
\end{array}
\]

There are other ones like this as well: cube and cube roots, raise to the fourth power and fourth roots, etc.

Another less well known one, and one that confuses many of my students, is exponential and logarithm operations.

When x is the base and a number, say 2, is in the exponent, as in x², this is a power operation. However, when x is the exponent, and a number called e is in the base, this is called an exponential operation, .

I won’t spend much time on e, but it is an irrational number like 𝜋 and there are several reasons why it is the ‘base of choice’. So if the operation is to take a number and make it the power of e, you will generate a number. Many calculators have an ‘.’ key where you can do this. So for example, e² = 7.389 … . What inverse operation can I perform on this number to get the ‘2’ back? This operation is called the log operation.

Now logs can use any base but again, the base e is most commonly used in math. The notation for taking the log of a number is logₑx which is often shortened to ln x. The shorthand ‘ln’ stands for ‘natural log’ as e is a natural base to use in math (topic for another post). If you have an ‘ ‘ key on your calculator, you probably also have an ‘logₑx‘ or ‘ln x‘ key as well.

So what does ‘ln x‘ mean? This operation is asking the mathematical question: “What number do I need to raise e to, so I get x?”. Can you see how this is the opposite of ‘eˣ ‘ where I actually raise e to the x power? So, ln() is asking the maths question: “What power do I need to raise e to, so that I get ?”. Isn’t it obvious that x is the number I have to raise e to, to get ? So and ln x are inverse operations. If you take the number we generated before, 7.389 … , and hit the ln x key on your calculator, you will get the original ‘2’ back.

There are many other inverse functions. In trigonometry, the sin x key in your calculator will give the sine of the angle x. Sometimes 𝜃 is used instead of x. The inverse sine, sometimes called the arcsine, is the inverse of this. The notation you may see on your calculator for this operation is ‘ASIN x‘ or ‘SIN ̅¹x‘. SIN ̅¹x is asking the maths question: ” What angle has x as its sine?”. So SIN ̅¹[SIN(x)] is x.

There are inverse functions associated with the other trig functions as well. You will usually see inverse function keys as the same key on a calculator, you just need to hit another key first to activate the inverse definition for the key.

Fractional Exponents Examples

In my last post, I ended with

\[
{x}^{\frac{m}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{{x}^{m}}\hspace{0.33em}{=}\hspace{0.33em}{\left({\sqrt[n]{x}}\right)}^{m}
\]

If I rewrite this, replacing the radical symbols with their exponent equivalents, I get

\[
{x}^{\frac{m}{n}}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}^{m}{)}^{\frac{1}{n}}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}^{\frac{1}{n}}{)}^{m}
\]. Note that \[
{m}\hspace{0.33em}\times\hspace{0.33em}\frac{1}{n}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{n}\hspace{0.33em}\times\hspace{0.33em}{m}\hspace{0.33em}{=}\hspace{0.33em}\frac{m}{n}
\].

This seems to imply that if you have a base raised to a power and that is again raised to a power, you can simplify this by just multiplying the exponents. This is, in fact, true. You can show this for yourself by expanding

\[
{(}{2}^{3}{)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}^{{3}\times{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{6}
\].

In general,

\[
{(}{x}^{m}{)}^{n}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{m}\times{n}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{mn}
\].

By the way, \[
{x}^{0}\hspace{0.33em}{=}\hspace{0.33em}{1}
\] no matter how small or large x is, except for x = 0. \[
{0}^{0}
\] is not defined and if you try to do it, like dividing by 0, the math police will be knocking on your door.

Two other exponent rules, which you can demonstrate for yourself with simple examples are:

\[
{(}{xy}{)}^{m}\hspace{0.33em}{=}\hspace{0.33em}{x}^{m}{y}^{m}{,}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\left({\frac{x}{y}}\right)}^{m}\hspace{0.33em}{=}\hspace{0.33em}\frac{{x}^{m}}{{y}^{m}}
\]

With these rules and the ones covered previously, let’s do some examples:

  1. \[
    {(}{x}^{2}{y}{)(}{x}^{3}{y}^{4}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{2}{+}{3}}{y}^{{1}{+}{4}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{5}{y}^{5}
    \]
  2. \[
    \frac{{x}^{2}{y}^{5}}{{xy}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{2}{-}{1}}{y}^{{5}{-}{3}}\hspace{0.33em}{=}\hspace{0.33em}{xy}^{2}
    \]
  3. \[
    {\left({7{a}^{3}{b}^{{-}{1}}}\right)}^{0}\hspace{0.33em}{=}\hspace{0.33em}{1}
    \]
  4. \[
    {\left({{z}^{5}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{z}^{{5}\times{2}}\hspace{0.33em}{=}\hspace{0.33em}{z}^{10}
    \]
  5. \[
    {\left({\frac{2x}{3{y}^{2}}}\right)}^{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}^{3}{x}^{3}}{{3}^{3}{y}^{{2}\times{3}}}\hspace{0.33em}{=}\hspace{0.33em}\frac{8{x}^{3}}{27{y}^{6}}
    \]  (This problem uses several rules at the same time)
  6. \[
    {\left({{-}{2}{x}^{2}{y}^{{-}{4}}}\right)}^{{-}{2}}\hspace{0.33em}{=}\hspace{0.33em}\left({{-}{2}^{{-}{2}}{x}^{{2}\times{(}{-}{2}{)}}{y}^{{-}{4}\times{(}{-}{2}{)}}}\right)\hspace{0.33em}{=}\hspace{0.33em}\left({{-}{2}^{{-}{2}}{x}^{{-}{4}}{y}^{8}}\right)\hspace{0.33em}{=}\hspace{0.33em}\frac{{y}^{8}}{4{x}^{4}}
    \]

If you have questions regarding any of these, send me an email via the Contact page, leave a comment via the Comments page, or post a comment on my facebook page.

Exponents and Roots

Well you might suspect that there is a connection between the roots of a number and exponents given my last couple of posts, and you would be correct. Now in my post on roots, I limited myself to using numbers that are perfect roots. For example, \[
\sqrt[3]{8}\hspace{0.33em}{=}\hspace{0.33em}{2}
\]. But many roots do not have such a simple answer. You just have to represent it exactly as a root. For example, \[
\sqrt{2}
\] has no simple integer solution. In fact, its decimal equivalent cannot be written down exactly as it is a non-repeating decimal. The answer you get on a calculator is just the first few decimals. \[
\sqrt{2}\] is an example of something called an irrational number. That doesn’t mean you can’t reason with it. It just means you can’t write it down exactly using decimals. You can only exactly represent it as \[
\sqrt{2}
\].

But it is still true that \[
\sqrt{2}\hspace{0.33em}\times\hspace{0.33em}\sqrt{2}\hspace{0.33em}{=}\hspace{0.33em}{2}
\] (ignoring the negative solution for now).

Now from my post on exponents, you saw that

\[
{2}^{3}\hspace{0.33em}\times\hspace{0.33em}{2}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}^{{3}{+}{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{5}
\]

Now bear with me, but consider

\[
{2}^{\frac{1}{2}}\hspace{0.33em}\times\hspace{0.33em}{2}^{\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{\frac{1}{2}{+}\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{1}\hspace{0.33em}{=}\hspace{0.33em}{2}
\].

But remember that

\[
\sqrt{2}\hspace{0.33em}\times\hspace{0.33em}\sqrt{2}\hspace{0.33em}{=}\hspace{0.33em}{2}
\].

Perhaps

\[
{2}^{\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{2}
\].

This is in fact true and the rules of exponents apply here as well. In general,

\[
{x}^{\frac{1}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{x}
\].

Now consider

\[
\sqrt{2}\hspace{0.33em}\times\hspace{0.33em}\sqrt{2}\hspace{0.33em}\times\hspace{0.33em}\sqrt{2}\hspace{0.33em}{=}\hspace{0.33em}{2}^{\frac{1}{2}}\hspace{0.33em}\times\hspace{0.33em}{2}^{\frac{1}{2}}\hspace{0.33em}\times\hspace{0.33em}{2}^{\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{\frac{3}{2}}
\].

Looks like a number raised to a fraction does have meaning. Now in decimals, this number is 2.82842712475… , which actually goes on forever. But on a calculator, you can get this answer by first raising 2 to the “3” power then taking the square root of the result, or first take the square root of 2 then raise the result to the “3” power. In general:

\[
{x}^{\frac{m}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{{x}^{m}}\hspace{0.33em}{=}\hspace{0.33em}{\left({\sqrt[n]{x}}\right)}^{m}
\].

In my next post, I will show several examples of fractional exponents.

Exponent Rules

So I will get back to roots of numbers, but let’s first look at the rules for combining exponents.

Now an integer exponent means that you are multiplying the base together as many times as the exponent indicates. That is,

\[
{3}^{5}\hspace{0.33em}{=}\hspace{0.33em}{3}\hspace{0.33em}\times\hspace{0.33em}{3}\hspace{0.33em}\times\hspace{0.33em}{3}\hspace{0.33em}\times\hspace{0.33em}{3}\hspace{0.33em}\times\hspace{0.33em}{3}
\].

The exponent “5” says to multiply the base “3” five times. This immediately suggests our first rule of exponents:

\[
{x}^{m}{x}^{n}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{m}{+}{n}}
\]

That is, when the same base with exponents are multiplied together, you can simplify this by adding the exponents. You can readily see this with the example above:

\[
{3}^{2}\hspace{0.33em}\times\hspace{0.33em}{3}^{3}\hspace{0.33em}{=}\hspace{0.33em}{(}{3}\hspace{0.33em}\times\hspace{0.33em}{3}{)}\hspace{0.33em}\times\hspace{0.33em}{(}{3}\hspace{0.33em}\times\hspace{0.33em}{3}\hspace{0.33em}\times\hspace{0.33em}{3}{)}\hspace{0.33em}{=}\hspace{0.33em}{3}^{{2}{+}{3}}\hspace{0.33em}{=}\hspace{0.33em}{3}^{5}
\]

So this rule makes sense. Now let’s look at another example to motivate the next exponent rule:

\[
\frac{{3}^{3}}{{3}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}{3}}{\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{1}\hspace{0.33em}{=}\hspace{0.33em}{3}^{1}
\]

Now you would normally leave out the exponent “1” in the final answer but I left it there so you can see the following rule in action:

\[
\frac{{x}^{m}}{{x}^{n}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{m}{-}{n}}
\]

Can you see how this rule works for the last example? By the way, these rules work whether or not the base is a known number or not.

\[
{x}^{13}{x}^{7}\hspace{0.33em}{=}\hspace{0.33em}{x}^{20}{,}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\frac{{x}^{13}}{{x}^{7}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{6}
\]

 

Now so far, I have limited myself to using positive integers as the exponents. It turns out that any number can be used as an exponent but it is not clear what a negative or non-integer exponent means. Let’s first look at negative integer exponents.

In the division example above with the base “3”, I specifically put the “3” with the larger exponent in the numerator. What if I reversed these:

\[
\frac{{3}^{2}}{{3}^{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}}{\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{3}^{1}}
\]

But according to the division rule of exponents:

\[
\frac{{3}^{2}}{{3}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{3}^{{2}{-}{3}}\hspace{0.33em}{=}\hspace{0.33em}{3}^{{-}{1}}
\]

This suggests that \[
{3}^{{-}{1}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{3}^{1}}
\] and this is correct. A negative exponent of a base means the equivalent to the same base to the positive exponent in the denominator. It actually works in the other direction as well. You can move factors between the numerator and denominator as long as you change the sign of the exponent:

\[
\begin{array}{l}
{{x}^{{-}{6}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{x}^{6}}}\\
{\frac{{x}^{{-}{6}}\hspace{0.33em}{y}^{5}}{{z}^{{-}{7}}}\hspace{0.33em}{=}\hspace{0.33em}\frac{{y}^{5}{z}^{7}}{{x}^{6}}}
\end{array}
\]

And the multiplication rule works as well:

\[
{x}^{7}{x}^{{-}{4}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{7}{-}{4}}\hspace{0.33em}{x}^{3}
\]

That takes care of the numbers on the tick marks of the number line as exponents, but what about the numbers in between? That will be the topic of my next post.

Equations with Exponents

So let see how to solve \[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{18}
\]. A good first step would be to remove the -7 from the left side. We do this by adding +7:

\[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{18}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{+}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{18}\hspace{0.33em}{+}\hspace{0.33em}{7}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{x}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}
\]

 

Now before I proceed further, after doing things like adding or subtracting a number to both sides of an equation, you start to notice mental shortcuts. For example, adding the 7 to both sides of the equation, effectively is the same as moving the 7 from the left side to the right side and changing the sign. This can be done with any term: move it to the other side of the equation and change the sign. By the way, a term is anything that is added or subtracted to or from the entire side of an equation. I’ll illustrate this further in future equations.

So now we have \[{x}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}
\]. So how do we get x by itself? From yesterday’s post, we know that:

\[
\sqrt{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{x}
\]

So this suggests that we need to take the square root of both sides of the equation:

\[
{x}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\sqrt{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[]{25}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}\pm\hspace{0.33em}{5}
\]

 

Did you remember that taking the square root results in two solutions? If you replace the x in the original equation with either +5 or -5, you will get a true equation 18 = 18.

Now many time the equation comes from a physical problem, like solving for a length. If so, then you can ignore the negative solution as that would not make sense. There are times, however, when the negative solution is the desired one.

More on Exponents

In my last post, you saw another maths shortcut – exponents. But I used examples where the exponents were positive integers. It turns out that mathematically, any number, be it negative, fractional, or irrational, can be used as an exponent. Let’s expand our knowledge here by looking at exponents that are not positive integers.

In my last post, I presented the rule that

\[
\frac{{x}^{a}}{{x}^{b}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{a}{-}{b}}
\]

so that

\[
\frac{{x}^{3}}{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{3}{-}{2}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{1}\hspace{0.33em}{=}\hspace{0.33em}{x}
\].

What if you have \[
\frac{{x}^{3}}{{x}^{3}}
\]?

Well, the rule says that

\[
\frac{{x}^{3}}{{x}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{3}{-}{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{0}
\]

But we know that the same number divided by itself is 1. So it makes sense, and it makes maths consistent, if we define anything raised to the “0” power as 1. That is,

\[
{x}^{0}\hspace{0.33em}{=}\hspace{0.33em}{1}
\] for any number x.

Now let’s go further. What if we have \[
\frac{{x}^{2}}{{x}^{3}}
\]? According to the rule, this is \[
{x}^{{2}{-}{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{-}{1}}
\]. What does a negative exponent mean? Well let’s do the same problem without using the rule:

\[
\frac{{x}^{2}}{{x}^{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{x}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{x}}{\rlap{/}{x}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{x}\hspace{0.33em}\times\hspace{0.33em}{x}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{x}^{1}}
\]

 

So it appears that a negative exponent means that the number raised to a negative power, is the same as the number in the denominator raised to its positive power. This is true for any exponent:

\[
{x}^{{-}{a}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{x}^{a}}
\]

 

This means that factors in a fraction can be moved at will from the numerator to the denominator or vice versa, by just changing the sign of the exponents:

\[
\begin{array}{l}
{\frac{7}{{x}^{{-}{2}}}\hspace{0.33em}{=}\hspace{0.33em}{7}{x}^{2}}\\
{{3}{y}^{{-}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{{y}^{3}}}\\
{\frac{4{xy}^{5}}{{wz}^{{-}{6}}}\hspace{0.33em}{=}\hspace{0.33em}\frac{4{xy}^{5}{z}^{6}}{w}}
\end{array}
\]

 

However, be careful. You can only do this with factors, that is things that are multiplied together. It does not work with fractions where things are added or subtracted:

\[
\frac{x}{{y}\hspace{0.33em}{+}\hspace{0.33em}{z}^{{-}{3}}}\hspace{0.33em}\ne\hspace{0.33em}\frac{{x}\hspace{0.33em}{+}\hspace{0.33em}{z}^{3}}{y}
\]

 

By the way, the symbol ≠ means “does not equal”.

More on Exponents

So now we know that \[
{x}^{3}
\] is shorthand math notation for x × x × x. In exponents, the x is called the base and the 3 is called the exponent or the power of x. Anything to the power of 2 is referred to as squared and anything to the power of 3 is referred to as cubed. These terms come from the formulas for finding the areas of a square and the volume of a cube.

There are several rules involving exponents that allow us to simplify expressions. So let’s look at

\[
\begin{array}{c}
{{x}^{2}\hspace{0.33em}\times\hspace{0.33em}{x}^{3}}\\
{\Longrightarrow\hspace{0.33em}{x}^{2}\hspace{0.33em}\times\hspace{0.33em}{x}^{3}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}\hspace{0.33em}\times\hspace{0.33em}{x}{)(}{x}\hspace{0.33em}\times\hspace{0.33em}{x}\hspace{0.33em}\times\hspace{0.33em}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}^{5}}
\end{array}
\]

 

So it appears that you can get the final result simply by adding the exponents together, this is correct as long as the exponents apply to the same base. For any two numbers a and b:

\[
{x}^{a}\hspace{0.33em}\times\hspace{0.33em}{x}^{b}\hspace{0.33em}{=}\hspace{0.33em}{x}^{a}{x}^{b}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{a}{+}{b}}
\]

Now you remember that when you see x, there is an implied “1” in front of it. Well, when dealing with exponents, if there is no visible exponent, there is an implied “1” there as well since \[
{x}^{1}\hspace{0.33em}{=}\hspace{0.33em}{x}
\]. So what about a rule for division?

\[
\frac{{x}^{3}}{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{x}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{x}\hspace{0.33em}\times\hspace{0.33em}{x}}{\rlap{/}{x}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{x}}\hspace{0.33em}{=}\hspace{0.33em}\frac{x}{1}\hspace{0.33em}{=}\hspace{0.33em}{x}^{1}\hspace{0.33em}{=}\hspace{0.33em}{x}
\]

 

Because of the common factors of x in the numerator and denominator, I can cancel these away, leaving “1” in their place. So it appears that to get the final result, you just subtract the exponent in the denominator from the one in the numerator. Again, this is correct as long as the base is the same:

\[
\frac{{x}^{a}}{{x}^{b}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{a}{-}{b}}
\]

 

There are a few more things about exponents I want to cover before we use these in equations. This will be covered in my next post.

Exponents

Now let’s add another maths symbol: exponents. Now multiplication is really successive adding: 2 × 3 = 2 + 2 + 2 or 3 + 3. In other words, 2 × 3  can be thought of as adding 2 three times or adding 3 two times. Well exponents are indicating successive multiplication. Example:

\[
\begin{array}{l}
{{3}^{2}\hspace{0.33em}{=}\hspace{0.33em}{3}\hspace{0.33em}\times\hspace{0.33em}{3}}\\
{{2}^{3}\hspace{0.33em}{=}\hspace{0.33em}{2}\hspace{0.33em}\times\hspace{0.33em}{2}\hspace{0.33em}\times\hspace{0.33em}{2}}\\
{{x}^{3}\hspace{0.33em}{=}\hspace{0.33em}{x}\hspace{0.33em}\times\hspace{0.33em}{x}\hspace{0.33em}\times\hspace{0.33em}{x}}
\end{array}
\]

I will expand on this tomorrow.