## Vestiges of the Past Making Math Confusing

Something in Math HAS to Change

Convention is a beautiful thing.  It allows us to use symbols to convey little things like direction or a sound.  We can piece those things together to make larger things, and eventually use it to create something like what you’re reading now.  There are no inherent meanings to these shapes we call letters, or the sounds we use when speaking.  It all works because we agree, somehow, upon what they mean.  Of course, over generations and cultures, and between even different languages, some things get crossed up in translation, but it’s still pretty powerful.

The structure of writing, punctuation, and the Oxford comma, they all work because we agree.  We can look back and try to see the history of how the conventions have changed and sometimes find interesting connections.  Sometimes, there are artifacts from our past that just don’t really make sense anymore.  Either the language has evolved passed the usefulness, or the language adopted other conventions that conflict.

One example of this is the difference between its and it’s.  An apostrophe can be used in a conjunction and can also be used to show ownership.  Pretty simple rule to keep straight with its and it’s, but whose and who’s.  Why is it whose, with an e at the end?

According to my friendly neighborhood English teacher there was a great vowel shift, which can be read about here, where basically, people in around the 15th century wanted to sound fancy and wanted their words to look fancy when written.  So the letters e and b were added to words like whose and thumb.

Maybe we should take this one step further, and use thumbe.  Sounds good, right?

But then, there’s the old rule, i before e except after c, except in words like neighbor and weight, and in the month of May, or on a Tuesday.  Weird, er, wierd, right?

All said, not a big deal because those tricks of language will not cause a student to be illiterate.  A student can mix those things up and still have access to symbolism and writing and higher level understanding of language.

There are some conventions in math that work this way, too.  There are things that simply are a hold-over of how things were done a long time ago.  The convention carries with it a history, that’s what makes it powerful.  But sometimes the convention needs to change because it no longer is useful at helping making clear the intentions of the author.

One of the issues with changing this convention is that the people who would be able to make such changes are so well versed in the topic, they don’t see it as an issue.  Or, maybe they do, but they believe that since they got it right, figured it out, so could anybody else.

There is one particular thing in math that stands out as particularly problematic.  The radical symbol, it must go!  There’s a much more elegant method of writing that is intuitive and makes sense because it ties into other, already established ways of writing mathematics.

But, before I get into that exactly, let me say there’s an ancillary issue at hand. It starts somewhere in 3rd or 4th grade here in the US and causes problems that are manifested all the way through Calculus.  Yup, it’s multiplication.

Let me take just a moment to reframe multiplication by whole numbers and then by fractions for you so that the connection between those things and rational exponents will be more clear.

Consider first, 3 × 5, which is of course 15.  But this means we start with a group that has three and add it to itself five times.

Much like exponents are repeated multiplication, multiplication is repeated addition.  A key idea here is that with both we are using the same number over and again, the number written first.  The second number describes how many times we are using that first number.

Now of course 3 × 5 is the same as 5 × 3, but that doesn’t change the meaning of the grouping as I described.

3 + 3 + 3 + 3 + 3 = 3 × 5

Now let’s consider how this works with a fraction.

15 × ⅕.  The denominator describes how many times a number has been added to itself to arrive at fifteen.  We know that’s three.  So 15 × ⅕ = 3.

3 + 3 + 3 + 3 + 3 = 15

Three is added to itself five times to arrive at fifteen.

Let’s consider 15 × ⅖, where the five in the denominator is saying we are looking for a number that’s been repeatedly added to get to 15, but exactly added to itself 5 times.

In other words, what number can you add to itself to arrive at 15 in five equal steps?  That’s ⅕.

The two in the numerator is asking, how far are you after the 2nd step?

3 + 3 + 3 + 3 + 3 = 15

The second step is six.

Another way to see this is shown below:

3 →6→9→12→15

Step 1: 3 → Step 2: 6 → Step 3: 9 → Step 4: 12 → Step 5: 15

Thinking of it this way we can easily see that 15 × ⅘ is 12 and 15 × 5/5 is 15.  All of this holds true and consistent with the other ways we thinking about fractions.

So we see how multiplication is repeated addition of the same number and how fractions ask questions about the number of repeats taken to arrive at an end result.

Exponents are very similar, except instead of repeated addition they are repeated multiplication.

Multiplication:  3 × 5 = 3 + 3 + 3 + 3 + 3

Exponents:  3⁵ = 3 × 3 × 3 × 3 × 3

Do you see how the trailing numbers describe how many of the previous number there exists, but the way the trailing number is written, as normal text or a superscript (tiny little number up above), informs the reader of the operation?

Pretty cool, eh?

Just FYI, 3 times itself 5 times is 243.

15 × ⅕ = 3, because 3 + 3 + 3 + 3 + 3 = 15.  That is, three plus itself five times is fifteen.

2431/5= 3 because 3 × 3 × 3 × 3 × 3 = 243.  That is, three times itself five times is two hundred and forty three.

You might be thinking, big deal... but watch how much simpler this way of thinking about rational exponents is with something like an exponent of ⅗.  Let’s look at this like steps:

3 × 3 × 3 × 3 × 3 = 243

3→9→27→81→243

Step one is three, step two is nine, step three is twenty-seven, the fourth step is eighty one, and the fifth step is 243.  So, 2433/5is asking, looking at the denominator first, what number multiplied by itself five times is 243, and the numerator says, what’s the third step?  Twenty-seven, do you see?

Connecting the notation this way makes it simple and easy to read.  The only tricky parts would be the multiplication facts.

## Where the First “Law” of Logarithms Originates – Wednesday’s Why E6

rule 1

Wednesday’s Why $–$ Episode 6

In this week’s episode of Wednesday’s Why we will tackle why the following is “law” of exponents is true in a way that hopefully will promote mathematical fluency and confidence. It is my hope that through these Wednesday’s Why episodes that you are empowered to seek deeper understanding by seeing that math is a written language and that by substituting equivalent expressions we can manipulate things to find truths.

${\mathrm{log}}_{2}\left(M\cdot N\right)={\mathrm{log}}_{2}M+{\mathrm{log}}_{2}N$

Now of course the base of 2 is arbitrary, but we will use a base of two to explore this.

The first thing to be aware of is that exponents and logarithms deal with the same issue of repeated multiplication. There connection between the properties of each are tightly related. What we will see here is that the property of logarithms above and this property of exponents below are both at play here. But it is not so easy to see, so let’s do a little exploration.

Just to be sure of how exponents and logarithms are written with the same meaning, consider the following.

${a}^{b}=c↔{\mathrm{log}}_{a}c=b$

Let us begin with statement 1: ${2}^{A}=M$

We can rewrite this as a logarithm, statement 1.1: ${\mathrm{log}}_{2}M=A$

Statement 2:
${2}^{B}=N$

We can rewrite this as a logarithm, statement 2.1: ${\mathrm{log}}_{2}N=B$

If we take the product of M and N, we would get
2A ·2B. Since exponents are repeated multiplication,

2A ·2B = 2A + B

This gives us statement 3: $M\cdot N={2}^{A+B}$

Let us rewrite statement 3 as a logarithmic equation.

$M\cdot N={2}^{A+B}\to {\mathrm{log}}_{2}\left(M\cdot N\right)=A+B$

In statements 1.1 and 2.1 we see what A and B equal. So let’s substitute those now.

$\begin{array}{l}{\mathrm{log}}_{2}\left(M\cdot N\right)=A+B\\ {\mathrm{log}}_{2}\left(M\cdot N\right)={\mathrm{log}}_{2}M+{\mathrm{log}}_{2}N\end{array}$

It took a little algebraic-juggling to get it done, but hopefully you can now see that this is not a law or a rule, but a property of repeated multiplication, just like all of the properties of exponents are consequences of repeated multiplication.

Let me know what worked for you here and what did not. Leave me a comment.

Thank you again.

## Exponents Part 2

two

Exponents Part 2

Division

In the previous section we learned that exponents are repeated multiplication, which on its own is not tricky. What makes exponents tricky is determining what is a base and what is not for a given exponent. It is imperative that you really understand the material from the previous section before tackling what’s next. If you did not attempt the practice problems, you need to. Also watch the video that review them.

In this section we are going to see why anything to the power of zero is one and how to handle negative exponents, and why they mean division.

What Happens with Division and Exponents?

Consider the following expression, keeping in mind that the base is arbitrary, could be any number (except zero, which will be explained soon).

${3}^{5}$

This equals three times itself five total times:

${3}^{5}=3\cdot 3\cdot 3\cdot 3\cdot 3$

Now let’s divide this by 3. Note that 3 is just 31.

$\frac{{3}^{5}}{{3}^{1}}$

If we write this out to seek a pattern that we can use for a short-cut, we see the following:

$\frac{{3}^{5}}{{3}^{1}}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{3}$

If you recall how we explored reducing Algebraic Fractions, the order of division and multiplication can be rearranged, provided the division is written as multiplication of the reciprocal. That is how division is written here.

$\frac{{3}^{5}}{{3}^{1}}=\frac{3}{3}\cdot \frac{3\cdot 3\cdot 3\cdot 3}{1}$

And of course 3/3 is 1, so this reduces to:

$\frac{{3}^{5}}{{3}^{1}}=3\cdot 3\cdot 3\cdot 3={3}^{4}$

The short-cut is:

$\frac{{3}^{5}}{{3}^{1}}={3}^{5-1}={3}^{4}$

That is, if the bases are the same you can reduce. Reducing eliminates one of the bases that is being multiplied by itself from both the numerator and the denominator. A general form of the third short-cut is here:

Short-Cut 3: $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$

This might seem like a worthless observation, but this will help articulate the very issue that is going to cause trouble with exponents and division.

$\frac{{3}^{5}}{{3}^{1}}={3}^{5}÷{3}^{1}$ .

But that is different than

${3}^{1}÷{3}^{5}$

The expression above is the same as

$\frac{{3}^{1}}{{3}^{5}}$

This comes into play because

$\frac{{3}^{1}}{{3}^{5}}={3}^{1-5}$,

and 1 $–$ 5 = -4.

Negative Exponents?

In one sense, negative means opposite. Exponents mean multiplication, so a negative exponent is repeated division. This is absolutely true, but sometimes difficult to write out. Division is not as easy to write as multiplication.

Consider that 3-4 is 1 divided by 3, four times. 1 ÷ 3 ÷ 3 ÷ 3 ÷ 3. But if we rewrite each of those ÷ 3 as multiplication by the reciprocal (1/3), it’s must cleaner and what happens with a negative exponent is easier to see.

$1÷3÷3÷3÷3\to 1\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}$

This is classically repeated multiplication. While one times itself any number of times is still one, let’s go ahead and write it out this time.

$1\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\to 1\cdot {\left(\frac{1}{3}\right)}^{4}$

This could also be written:

$1\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\to 1\cdot \frac{{1}^{4}}{{3}^{4}}$

The second expression is easier, but both are shown here to make sure you see they are the same.

Since 1 times 14 is just one, we can simplify this further to:

$1\cdot \frac{{1}^{4}}{{3}^{4}}=\frac{1}{{3}^{4}}.$

Negative exponents are repeated division. Since division is hard to write and manipulate, we will write negative exponents as multiplication of the reciprocal. In fact, if instructions say to simplify, you cannot have a negative exponent in your final answer. You must rewrite it as multiplication of the reciprocal. Sometimes that can get ugly. Consider the following:

$\frac{b}{{a}^{-5}}$

To keep this clean, let us consider separating this single fraction as the product of two rational expressions.

$\frac{b}{{a}^{-5}}=\frac{b}{1}\cdot \frac{1}{{a}^{-5}}$

The b is not a problem here, but the other rational expression is problematic. We need to multiply by the reciprocal of $\frac{1}{{a}^{-5}}$, which is just a5.

$\frac{b}{{a}^{-5}}=\frac{b}{1}\cdot \frac{{a}^{5}}{1}={a}^{5}b$.

This can also be considered a complex fraction, the likes of which we will see very soon. Let’s see how that works.

$\frac{b}{{a}^{-5}}$

Note: ${a}^{-5}=\frac{1}{{a}^{5}}$

Substituting this we get:

$\frac{b}{\frac{1}{{a}^{5}}}$

This is b divided by 1/a5.

$b÷\frac{1}{{a}^{5}}$

Let’s multiply by the reciprocal:

$b\cdot {a}^{5}$

Now we will rewrite it in alphabetical order (a good habit, for sure).

${a}^{5}b$

Let us consider one more example before we make our fourth short-cut. With this example we could actually apply our second short-cut, but it will not offer much insight into how these exponents work with division.

This is the trickiest of all of the ways in which exponents are manipulated, so it is worth the extra exploration.

$\frac{2{x}^{-2}{y}^{-5}z}{{2}^{-2}x{y}^{3}{z}^{-5}}$

As you see we have four separate bases. In order to simplify this expression we need one of each base (2, x, y, z), and all positive exponents. So let’s separate this into the product of four rational expressions, then simplify each.

$\frac{2{x}^{-2}{y}^{-5}z}{{2}^{-2}x{y}^{3}{z}^{-5}}\to \frac{2}{{2}^{-2}}\cdot \frac{{x}^{-2}}{x}\cdot \frac{{y}^{-5}}{{y}^{3}}\cdot \frac{z}{{z}^{-5}}$

The base of two first:

$\frac{2}{{2}^{-2}}\to 2÷{2}^{-2}$

We wrote it as division. What we will see is dividing is multiplication by the reciprocal, and then the negative exponent is also dividing, which is multiplication by the reciprocal. The reciprocal of the reciprocal is just the original. But watch what happens with the sign of the exponent.

First we will rewrite the negative exponent as repeated division.

$2÷\frac{1}{{2}^{2}}$

Now we will rewrite division as multiplication by the reciprocal.

$2\cdot {2}^{2}={2}^{3}$

Keep in mind, this is the same as 23/1.

We will offer similar treatment to the other bases.

Consider first $\frac{{x}^{-2}}{x}=\frac{{x}^{-2}}{1}\cdot \frac{1}{x}$

Negative exponents are division, so:

$\frac{{x}^{-2}}{x}=\frac{{x}^{-2}}{1}\cdot \frac{1}{x}$

Notice the x that is already dividing (in the denominator) does not change. It has a positive exponent, which means it is already written as division.

$\frac{{x}^{-2}}{1}\cdot \frac{1}{x}\to \frac{1}{{x}^{2}}\cdot \frac{1}{x}=\frac{1}{{x}^{3}}$

This is exactly how simplifying the y and z will operation.

$\frac{{2}^{3}}{1}\cdot \frac{1}{{x}^{2}\cdot x}\cdot \frac{1}{{y}^{5}\cdot {y}^{3}}\cdot \frac{z\cdot {z}^{5}}{1}$

Putting it all together:

$\frac{2{x}^{-2}{y}^{-5}z}{{2}^{-2}x{y}^{3}{z}^{-5}}=\frac{{2}^{3}{z}^{6}}{{x}^{3}{y}^{8}}$.

Short-Cut 4: Negative exponents are division, so they need to be rewritten as multiplication by writing the reciprocal and changing the sign of the exponent. The last common question is what happens to the negative sign for the reciprocal? What happens to the division sign here: $3÷5=3×\frac{1}{5}$. When you rewrite division you are writing it as multiplication. Positive exponents are repeated multiplication.

${a}^{-m}=\frac{1}{{a}^{m}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{{a}^{-m}}={a}^{m}$

This is the second to last thing we need to learn about exponents. However, a lot of practice is required to master them fully.

To see why anything to the power of zero is one, let’s consider:

${3}^{5}$

This equals three times itself five total times:

${3}^{5}=3"#x22C5;3\cdot 3\cdot 3\cdot 3$

Now let’s divide this by 35.

$\frac{{3}^{5}}{{3}^{5}}$

Without using short-cut 3, we have this:

$\frac{{3}^{5}}{{3}^{5}}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3}=1$

Using short-cut 3, we have this:

$\frac{{3}^{5}}{{3}^{5}}={3}^{5-5}$

Five minutes five is zero:

${3}^{5-5}={3}^{0}$

Then 30 = 1.

Τhe 3 was an arbitrary base. This would work with any number except zero. You cannot divide by zero, it does not give us a number.

The beautiful thing about this is that no matter how ugly the base is, if the exponent is zero, the answer is just one. No need to simplify or perform calculation.

${\left(\frac{{3}^{2x-1}\cdot {e}^{\pi i}}{\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}}\right)}^{0}=1$

Let’s take a quick look at all of our rules so far.

 Short-Cut Example ${a}^{m}\cdot {a}^{n}={a}^{m+n}$ ${5}^{8}\cdot 5={5}^{8+1}={5}^{9}$ ${\left({a}^{m}\right)}^{n}={a}^{mn}$ ${\left({7}^{2}\right)}^{5}={7}^{10}$ $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ $\frac{{5}^{7}}{{5}^{2}}={5}^{7-2}={5}^{5}$ ${4}^{-3}=\frac{1}{{4}^{3}}$ ${a}^{0}=1$ 50 = 1

Let’s try some practice problems.

Instructions: Simplify the following.

1. ${\left({2}^{8}\right)}^{1/3}$ 2. $3{x}^{2}\cdot {\left(3{x}^{2}\right)}^{3}$

3. $\frac{5}{{5}^{m}}$ 4. $\frac{{5}^{2}{x}^{-3}{y}^{5}}{{5}^{-3}{x}^{-4}{y}^{-5}}$

5. $7÷7÷7÷7÷7÷7÷7÷7$ 6. $9{x}^{2}y÷9{x}^{2}y$

7. $9{x}^{2}y÷\left(9{x}^{2}y\right)$ 8. ${\left({x}^{2}\cdot 2{x}^{6}\right)}^{2}\cdot {\left({x}^{2}\cdot 2{x}^{6}\right)}^{-2}$

9. ${\left({a}^{m}\right)}^{n}\cdot {a}^{m}$ 10. $\frac{{\left(3{x}^{2}+4\right)}^{2}}{{\left(3{x}^{2}+4\right)}^{3}}$

## Exponents Part 1

exponents part 1

Exponents Part 1

One of the biggest things to understand about math is how it is written. The spatial arrangement of characters is syntax. Syntax, in English, refers to the arrangements of words to convey meaning.

Exponents are just a way of writing repeated multiplication. If we are multiplying a number by itself repeatedly, we can use an exponent to tell how many times the number is being multiplied. That’s it. Nothing tricky exists with exponents, no new operations or concepts to tackle. If you’re familiar with multiplication and its properties, exponents should be accessible.

That said, it is not without its pitfalls. A balance between conceptual understanding and procedural short-cuts is needed to avoid those pitfalls. The only way to strike that balance is through a careful progression of exercises and examples. An answer-getting mentality will lead to big troubles with exponents. People wishing to learn how exponents work must seek understanding.

Let’s establish some facts that will come into play with this first part of exponents.

1.      Exponents are repeated multiplication

To simplify simple expressions with exponents you only need to know a few short-cuts, but to recall and understand, we need more. These facts are important.

With an exponential expression we have a base, the number being multiplied by itself, and the exponent, the small number on the top right of the base which describes how many times the base is being multiplied by itself.

${a}^{5}$

The number a is the base. We don’t know what a is other than it is a number. It’s not a big deal that we don’t know exactly what number it is, we still know things about this expression.

Five is the exponent, which means there are five a’s, all multiplying together, like this: $a\cdot a\cdot a\cdot a\cdot a.$

Something to keep in mind is that this expression equals another number. Since we don’t know what a is, we cannot find out exactly what it is, but we do know it’s a perfect 5th power number, like 32. See, 25 = 32.

What if we had another number multiplying with a5, like this:

${a}^{5}\cdot {b}^{3}$

If we write this out, without the exponents we see we have 5 a’s and 3 b’s, all multiplying together. We don’t know what a or b equals, but we do know they’re multiplying so we could change the order of multiplication (commutative property) or group them together in anyway we wish (associative property) without changing the value.

${a}^{5}\cdot {b}^{3}=a\cdot a\cdot a\cdot a\cdot a\cdot b\cdot b\cdot b$

And these would be the same:

$\left(a\cdot a\right)\left(a\cdot a\cdot a\right)\left(b\cdot b\cdot b\right)$

$\left(a\cdot a\right)\left[\left(a\cdot a\cdot a\right)\left(b\cdot b\cdot b\right)\right]$

$\left(a\cdot a\right){\left[ab\right]}^{3}$

${a}^{2}{\left[ab\right]}^{3}$

This is true because the brackets group together the a and b, making them both the base. The brackets put them together. The base is ab, and the exponent is 3. This means we have ab multiplied by itself three times.

Keep in mind, these are steps but exploring how exponents work to help you learn to read the math for the intended meaning behind the spatial arrangement of bases, parenthesis and exponents.

Now, the bracketed expression above is different than ab3, which is $a\cdot b\cdot b\cdot b$.

${\left(ab\right)}^{3}\ne a{b}^{3}$

Let’s expand these exponents and see why this is:

${\left(ab\right)}^{3}\ne a{b}^{3}$

Write out the base ab times itself three times:

$\left(ab\right)\left(ab\right)\left(ab\right)\ne a\cdot b\cdot b\cdot b$

The commutative property of multiplication allows us to rearrange the order in which we multiply the a’s and b’s.

$a\cdot a\cdot a\cdot b\cdot b\cdot b\ne a\cdot b\cdot b\cdot b$

Rewriting this repeated multiplication we get:

${a}^{3}{b}^{3}\ne a{b}^{3}$

The following, though, is true:

$\left(a{b}^{3}\right)=a{b}^{3}$

On the right, the a has only an exponent of 1. If you do not see an exponent written, it is one. If we write it out we see:

$\left(a\cdot b\cdot b\cdot b\right)=a\cdot b\cdot b\cdot b$

In summary of this first exploration, the base can be tricky to see. Parenthesis group things together. An exponent written outside the parenthesis creates all of the terms inside the parenthesis as the base. But if numbers are multiplying, but not grouped, and one has an exponent, the exponent only belongs to the number just below it on the left. For example, $4{x}^{3},$ the four has an exponent of just one, while the x is being cubed.

Consider: ${\left(x+5\right)}^{3}.$ This means the base is x + 5 and it is multiplied by itself three times.

Repeated Multiplication Allows Us Some Short-Cuts

Consider the expression:

${a}^{3}×{a}^{2}.$

If we wrote this out, we would have:

$a\cdot a\cdot a×a\cdot a$.

(Note: In math we don’t use colors to differentiate between two things. A red a and a blue a are the same. These are colored to help us keep of track of what’s happening with each part of the expression.)

This is three a’s multiplying with another two a’s. That means there are five a’s multiplying.

${a}^{3}×{a}^{2}={a}^{5}$

Before we generalize this to find the short-cut, let us see something similar, but is a potential pitfall.

${a}^{3}×{b}^{2}$

If we write this out we get:

$a\cdot a\cdot a×b\cdot b$

This would not be an exponent of 5, in anyway. An exponent of five means the base is being multiplied by itself five times. Here we have an a as a base, and three of those multiplying, and a b as a base, and two of those multiplying. Not five of anything.

The common language is that if the bases are the same we can add the exponents. This is a hand short-cut, but if you forget where it comes from and why it is true, you’ll undoubtedly confuse it with some of the other short-cuts that follow.

Short-Cut 1: If the bases are the same you can add the exponents. This is true because exponents are repeated multiplication and the associative property says that the order in which you group things does not matter (when multiplying).

${a}^{m}×{a}^{n}={a}^{m+n}$

The second short-cut comes from groups and exponents.

${\left({a}^{3}\right)}^{2}$

This means the base is a3, and it is being multiplied by itself.

${a}^{3}×{a}^{3}$

Our previous short cut said that if the bases are the same, we can add the exponents because we are just adding how many of the base is being multiplied by itself.

${a}^{3}×{a}^{3}={a}^{3+3}={a}^{6}$

But this is not much of a short cut. Let us look at the original expression and the outcome and look for a pattern.

${\left({a}^{3}\right)}^{2}={a}^{6}$

Short-Cut 2: A power raised to another is multiplied.

${\left({a}^{m}\right)}^{n}={a}^{m×n}$

Be careful here, though:

$a{\left({b}^{3}{c}^{2}\right)}^{5}$ = $a{b}^{15}{c}^{10}$

Practice Problems

 1.       ${x}^{4}\cdot {x}^{2}$ 8. ${\left(5xy\right)}^{3}$ 2.       ${y}^{9}\cdot y$ 9. ${\left(8{m}^{4}\right)}^{2}\cdot {m}^{3}$ 3.       ${z}^{2}\cdot z\cdot {z}^{3}$ 10. ${\left(3{x}^{5}\right)}^{3}{\left({3}^{2}{x}^{7}\right)}^{2}$ 4.       ${\left({x}^{5}\right)}^{2}$ 11. $7{\left({7}^{2}{x}^{4}\right)}^{5}\cdot {7}^{3}{x}^{5}$ 5.       ${\left({y}^{4}\right)}^{6}$ 12. ${5}^{3}+{5}^{3}+{5}^{3}+{5}^{3}+{5}^{3}$ 6.       ${x}^{3}+{x}^{3}+{x}^{3}+{x}^{8}+{x}^{8}$ 13. ${3}^{2}\cdot 9$ 7.       ${4}^{x}+{4}^{x}+{4}^{x}$ 14. ${4}^{x}\cdot {4}^{x}\cdot {4}^{x}$