## Newton’s Clock, Part 1

As I mentioned many posts before, maths is a language – a much more elegant and concise language when taking about maths stuff. This will be very apparent when I talk about the expressions in the clock I just bought for my office:

Each one of the maths expressions on the clock equal the number corresponding to its position on the clock. So I will be talking about each of these.

By far, the expression corresponding to “1” on the clock, illustrates the conciseness of maths expressions:

${e}^{\mathit{\pi}{i}}\cos\mathit{\pi}\hspace{0.33em}{=}\hspace{0.33em}{1}$

This expression makes use of more mathematical knowledge than any of the other expressions. It uses the maths constants e and 𝜋, and uses radians, trigonometric (or circular) functions, exponentials, and Euler’s identity. The two parts of this expression are e𝜋i and cos 𝜋. Let me first talk about cos 𝜋.

Now I have talked about trig functions before, but I mainly talked about sine of an angle (abbreviated “sin”). The cosine of an angle (abbreviated “cos”) is similar. You can take the cosine of an angle on your calculator, but you need to tell the calculator whether you are measuring angles in “degrees” or “radians”. In this expression, we are taking the cosine of 𝜋 radians. If your calculator is in radians mode and if you have ‘𝜋’ button, taking the cosine of 𝜋 will show -1. If you do not have a 𝜋 button, take the cosine of 3.14179 and you will get an answer approximately equal to -1. So, cos 𝜋 = -1.

Now the e𝜋i part requires a bit more of an explanation. The numbers e and 𝜋 have been discussed before. They are both irrational numbers which means that you cannot write them down exactly using numbers. We just agree that the symbol e is exactly e and the symbol 𝜋 is exactly 𝜋. If you have an ex button on your calculator, you can get an approximate value for e by typing 1, then the ex key. I have talked about e before. The letter e is for Euler who did a lot of work with this constant. So what is i?

The number i is not a real number. It is actually called an imaginary number. It is formally defined in terms of its square:

i2 = -1

which means that i is the square root of -1. I’m not making this up.

Since in the realm of real numbers, you know that you cannot take the square root of a negative number. So defining i is creating a whole new realm of numbers called complex numbers. Actually, complex numbers are the union of two realms: the reals and the imaginaries. There is a lot more to say about this but this will have to be a topic for a future post.

So what is e𝜋i where i is in the exponent? There is an identity (a maths rule) called Euler’s identity that explains what e𝜋i is equal to:

e𝜋i + 1 = 0

So e𝜋i must equal -1 for Euler’s identity to work. Again, explaining anything more is the topic of several future posts.

So if e𝜋i = -1 and cos 𝜋 = -1, then

e𝜋icos 𝜋 = (-1)(-1) = 1 which is the 1 o’clock spot on the clock.

I haven’t even scratched the surface of explaining this expression, but it has already filled up this post. The other expressions will not take as long to explain.

## Logarithms, Part 5

I never thought these posts would get to 5.

Now I said I would do a population problem but I have decided to go with a radioactive decay problem instead. I will use the example of carbon dating as this is based on radioactive decay. But first, let’s look at the general equation for exponential decay:

${A}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{kt}}$

This formula gives the amount of something that is decreasing exponentially. A is the amount left after t seconds, hours, days, years or whatever depending on the value of the rate of decrease factor which is k. A0 is the amount of something we started out with, the amount present at t = 0. This formula makes sense when you look at the ekt part of the equation.

Now I have talked about e before. It is an irrational number, like 𝜋, and is approximately equal to 2.7183. k is a rate of decrease factor that depends on the material we are working with and the units of time t. The –kt part, the exponent of e, is a negative number since k and t are positive. I have explained negative exponents before but ekt equals 1/ekt . Now what happens to ekt as t gets large? Any number greater than 1 (which e is) raised to a larger and larger power, gets very big. And when you divide a big number into 1, you get a very small number. So A0 is being multiplied by a number that gets smaller and smaller as time goes on. That is why A, the amount of material, is exponentially decreasing.

With that as a background, let’s talk about carbon dating. Any living thing has carbon in it. Indeed, all life on earth is carbon-based which means that the the molecules essential for life are composed of lots of carbon. Now carbon comes in different “flavors”. These flavors are called isotopes and carbon has two main isotopes: carbon 12 the most abundant and non-radioactive, and carbon 14 which is radioactive. Fortunately, the amount of carbon 14 is very small – about 1 atom to every 1012 atoms of carbon 12. However, in living things, this ratio is pretty much constant since carbon 14 is continually made in our atmosphere. But once something dies, the carbon 14 is not replenished and the amount present at the time of death starts decreasing.

So carbon dating is a process of determining the amount of carbon 14 left in a once living object then calculating the time it would take to have that much carbon 14 left.

So let’s go back to our equation for exponential decay. In order to use this equation for carbon dating, we need to know what k is for carbon 14. Now we know that the half-life of carbon 14 is 5700 years which means that given any amount of carbon 14, only half that amount will be left in 5700 years due to radioactive decay. So let’s use this fact to calculate k.

Taking this information and putting it into our equation results in

${0}{.}{5}{A}_{0}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{k}\times{5700}}$

So the left side shows that there is half (0.5) the initial amount and the right side shows that this occurs in 5700 years. So now, I will take the loge (abbreviated as ln) of both sides. Note that since e is the base on the right side, taking the log to that base just results in the exponent –kt. Also note that A0 appears on both sides of the equation, so we can divide both sides of the equation by A0 which makes A0 disappear:

$\begin{array}{l} {{0}{.}{5}{A}_{0}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{k}\times{5700}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{0}{.}{5}\hspace{0.33em}{=}\hspace{0.33em}{e}^{{-}{k}\times{5700}}}\\ {\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\ln{(}{0}{.}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}\ln{(}{e}^{{-}{k}\times{5700}}{)}\hspace{0.33em}{=}\hspace{0.33em}{-}{k}\times{5700}}\\ {\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{-}{0}{.}{6931}\hspace{0.33em}{=}\hspace{0.33em}{-}{k}\times{5700}}\\ {\Longrightarrow{k}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{0}{.}{6931}}{5700}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{0001216}} \end{array}$

So now that we know what k is, we can use the following equation to do our carbon dating:

${A}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{0}{.}{0001216}{t}}$

So let’s say a fossil has 35% (0.35) of its original carbon 14 when it died. How old is the fossil?

$\begin{array}{l} {{0}{.}{35}{A}_{0}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{0}{.}{0001216}{t}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{0}{.}{35}\hspace{0.33em}{=}\hspace{0.33em}{e}^{{-}{0}{.}{0001216}{t}}}\\ {\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\ln{(}{0}{.}{35}{)}\hspace{0.33em}{=}\hspace{0.33em}\ln{(}{e}^{{-}{0}{.}{0001216}{t}}{)}\hspace{0.33em}{=}\hspace{0.33em}{-}{0}{.}{0001216}{t}}\\ {\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{-}{1}{.}{0498}\hspace{0.33em}{=}\hspace{0.33em}{-}{0}{.}{0001216}{t}}\\ {\Longrightarrow{t}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{1}{.}{0498}}{{-}{0}{.}{0001216}}\hspace{0.33em}{=}\hspace{0.33em}{8633}\hspace{0.33em}{\mathrm{years}}} \end{array}$

We have a lot of birthdays to catch up on!

## Logarithms, Part 4

Let’s do another example using logarithms. As seen in my last post, logarithms are useful when the unknown variable in an equation is in the exponent of some number. But the exponent can be more than just the unknown – it can be an expression with an unknown. Consider the following problem:

${10}^{{3}{x}{+}{7}}\hspace{0.33em}{=}\hspace{0.33em}{125}$

So the first step, as seen last time, is to take the log of both sides of the equation. We then can use the property of logs that was introduced: ${\log}_{a}{b}^{x}\hspace{0.33em}{=}\hspace{0.33em}{x}\hspace{0.33em}{\log}_{a}{b}$

So let’s again use the base 10 log, the log x key on your calculator:

$\begin{array}{l} {{10}^{{3}{x}{+}{7}}\hspace{0.33em}{=}\hspace{0.33em}{125}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\log{(}{10}^{{3}{x}{+}{7}}{)}\hspace{0.33em}{=}\hspace{0.33em}\log{(}{125}{)}}\\ {\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{(}{3}{x}\hspace{0.33em}{+}\hspace{0.33em}{7}{)}\log{10}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{0969}} \end{array}$

Now let’s stop here. The log of 125 is done on your calculator. What about the log of 10? Well that can be done on your calculator as well, but if you’ve been paying attention, you can see that the answer is 1. 1 is the exponent of 10 to make 10? = 10. So now we have a standard (non-exponential) equation:

${3}{x}\hspace{0.33em}{+}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{0969}$

We have solved equations like this before, so without going into the detail, the solution to this is x = -1.6344. You can put this value of x in the left side of the original equation and find that it does solve it.

In my next post, I will present another property of logs and use it to solve a population problem.

## Logarithms, Part 3

Finally have a little time for a post.

So we know how to solve x2 = 10 by taking the square root of both side of the equation to get x = ±3.162… Note that taking the square root of x2 undoes or reverses the squaring of x.

But what do you do if x is in the exponent and not the base?

${2}^{x}\hspace{0.33em}{=}\hspace{0.33em}{10}$

You can’t take the xth root since you don’t know what x is. So what to do? From my last post, you saw that log2 10 means “what is the number that I can use as the exponent of 2 so that the answer is 10”. So in the above equation, if I take the log2 of both sides, I get

${\log}_{2}{2}^{x}\hspace{0.33em}{=}\hspace{0.33em}{\log}_{2}{10}$

The left side of this equation is doing two inverse operations on the number 2 – raising 2 to a power then taking its log. In other words, the left side can be seen as saying “what is the number that I can use as the exponent of 2 so that the answer is 2x ?”. Well the answer to that question is x. So the left side is just x and the right side is just a calculation:

${\log}_{2}{2}^{x}\hspace{0.33em}{=}\hspace{0.33em}{\log}_{2}{10}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{\log}_{2}{10}\hspace{0.33em}{=}\hspace{0.33em}{3}{.}{32192809488}$

Well that’s just dandy! Trouble is, without the internet, how do you find log2 10? I have not seen a calculator with a log2 x button. As mentioned in my last post, calculators usually have buttons to take logs relative to bases 10 and e. Well fortunately, there are lots of properties of logs that can help. The one we can use here is

${\log}_{a}{b}^{x}\hspace{0.33em}{=}\hspace{0.33em}{x}\hspace{0.33em}{\log}_{a}{b}$

This means that I can take the log with respect to any base, and the x can be removed as the exponent. So for our problem, let’s take the log10 (the log x key on your calculator) of both sides and see what happens:

${\log}_{10}{2}^{x}\hspace{0.33em}{=}\hspace{0.33em}{\log}_{10}{10}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}{\log}_{10}{2}\hspace{0.33em}{=}\hspace{0.33em}{1}$

Let’s stop here for a moment before I complete the solution. Why is the right side equal to 1? Log10 10 is saying “what power of 10 equals 10?”. The answer is 1 because 101 = 10. On the left side, I used to log property above to bring the x in front of the log. Now log10 2 is just a number. You can use the log x key on your calculator to find that log10 2 = 0.3010 to four decimal places. So now the equation becomes

${x}{\log}_{10}{2}\hspace{0.33em}{=}\hspace{0.33em}{1}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{0}{.}{301}{x}\hspace{0.33em}{=}\hspace{0.33em}{1}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{0.301}\hspace{0.33em}{=}\hspace{0.33em}{3}{.}{3219}$

So 23.3219 = 10. In my next post on logs, I’ll do more equation solving using logs.

## Logarithms, Part 2

So what is a logarithm? Let’s first see the notation, then I will explain. When taking the log (short for logarithm which I will use from now on) of a number, you need to know what base is being used. The notation for the log of x is loga x. The a is the base and is usually a specified number. so examples using this notation are log2 10, log10 25, log18 145, loge 7.34. Let’s look at these.

log2 10 is asking the question “What number can I use as the exponent of 2 so that the answer is 10?”. It turns out that 23.321928094887 = 10 so log2 10 = 3.321928094887.

log10 25 is asking the question “What number can I use as the exponent of 10 so that the answer is 25?”. Well, 101.39794 =25 so log10 25 = 1.39794.

Are you getting the picture? What about log18 145? This is asking the question “What number can I use as the exponent of 18 so that the answer is 145?”. 181.72183 = 145 so log18 145 = 1.72183.

Now let’s look at loge 7.34. This shows that the base or the number we are taking the log of does not have to be an integer. The number e, which I have talked about before, is an irrational number, but it still can be used as a base. In fact, it is probably the most used base. Since e1.99334 = 7.34, then it follows that loge 7.34 = 1.99334.

By the way, on most calculators, the log or log x key assumes that the base is 10. On most calculators as well, ln x means loge x. “ln” means “natural log”.

Now loga x and ax are inverses of each other. This means that one undoes the other. So if on your calculator, you find ln 7, then take that number and hit the ex key, you get the original 7 back. This works in reverse as well: Find e7 on your calculator, then hit the ln x key. You will again get the 7 back.

In notation-speak, this inverseness is shown as

$\begin{array}{l} {{a}^{{\log}_{a}x}\hspace{0.33em}{=}\hspace{0.33em}{x}}\\ {{\log}_{a}{a}^{x}{=}\hspace{0.33em}{x}} \end{array}$

In my next post, I will show how logarithms can be used to solve equations.

## Logarithms, Part 1

Logarithms confuse many of my students so I thought it is time to explain these. I touched on these before on a post about inverse operations, but let’s add some more detail.

Let’s first define some terms here. Consider the expression x2. Here, x is raised to the power of 2. x is the base and 2 is the exponent, power, order, or index. Lots of different terms for the exponent – I will mostly use the term exponent. So the exponent defines what to do with the base.

Now before I talk about logarithms specifically, I want to review what various kinds of exponents mean. I have talked about this before, but these concepts should be fully understood if logarithms are to make sense to you.

Now x2 means x × x. Positive integer exponents means how many times you multiply the base by itself. So in general, for a positive integer m,

xm = x × x × x × x … where x is listed m times.

The special case of when m = 0 is defined as x0 = 1, no matter how small or how large x is. Now what about negative integers?

$\begin{array}{c} {{x}^{{-}{1}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{x}^{1}}{;}\hspace{0.33em}\hspace{0.33em}{x}^{{-}{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{x}^{2}}{;}\hspace{0.33em}\hspace{0.33em}\frac{1}{{x}^{{-}{2}}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{2}}\\\ {{x}^{{-}{m}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{x}^{m}}{;}\hspace{0.33em}\hspace{0.33em}\frac{1}{{x}^{{-}{m}}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{m}} \end{array}$

So a negative exponent is the same as the positive one except it and its base is in the denominator or vice versa. You can freely move a factor that is a base and its exponent between the numerator and the denominator, as long as you change the sign of the exponent.

What about fractional exponents? Let’s start with fractions where “1” is in the numerator. The denominator in a fraction exponent refers to the root of the number. For example,

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

The “2” for the square root is usually assumed if it is not there. However, for other roots (like cube roots), the index must be there to indicate the kind of root it is. Other examples:

${x}^{\frac{1}{3}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[3]{x}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}^{\frac{1}{6}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[6]{x}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}^{\frac{1}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{x}$

The numerator in a fractional exponent means the same as if it wasn’t in a fraction. so we can combine these two definitions for more general fractions:

${x}^{\frac{2}{3}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[3]{{x}^{2}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}^{\frac{5}{6}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[6]{{x}^{5}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}^{\frac{m}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{{x}^{m}}$

Now we have not covered irrational exponents like x𝜋. The development of these are a bit more complex so I’ll just say “use your calculator”.

Indeed, you can use your calculator to calculate a number raised to a power if it has a key labelled as “yx” or has a key with the “^” symbol on it. I will leave it to you to find out how to use these keys. If you do not have a fancy calculator, there is always the all-knowing internet.

So we have talked before on how to solve equations like x2 = 16 by taking the square root of both sides of the equation. But how do you solve 2x = 16? Notice that x is now in the exponent. That changes everything as you can’t take the xth root of a number on your calculator……………but can you?

In the next post on this topic, I’ll introduce you to logarithms then later, how they are used.

## 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. ${\left({{-}{2}{x}^{2}{y}^{{-}{4}}}\right)}^{{-}{2}}\hspace{0.33em}{=}\hspace{0.33em}\left({{-}{2}^{{-}{2}}{x}^{{2}\times{(}{-}{2}{)}}{y}^{{-}{4}\times{(}{-}{2}{)}}}\right)\hspace{0.33em}{=}\hspace{0.33em}\left({{-}{2}^{{-}{2}}{x}^{{-}{4}}{y}^{8}}\right)\hspace{0.33em}{=}\hspace{0.33em}\frac{{y}^{8}}{4{x}^{4}}$

## 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.