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Lecture Comments (4)

1 answer

Last reply by: Professor Selhorst-Jones
Sat Mar 12, 2016 12:44 PM

Post by Jay Lee on March 12, 2016

When should we use the "absolute value" sign (for questions that ask us to change the expression from the root form to the exponential form)?

Thank you very much:)

1 answer

Last reply by: Professor Selhorst-Jones
Tue Aug 4, 2015 8:49 AM

Post by Duy Nguyen on August 4, 2015

Hi,suddenly I have unsupported format issue with educator.com
The video will not play. I am using a Mac. Please advise.
Thank you very much

Understanding Exponents

  • At heart, exponentiation is repeated multiplication. By definition, for any number x and any positive integer a,
    xa    =   

    x·x ·x …x ·x
    We can expand on this fundamental idea to see how exponentiation can work with numbers that aren't just positive integers.
  • Through multiplication, we can combine numbers that have the same exponent base:
    xa ·xb = xa+b.
  • We can consider exponentiation acting upon exponentiation:
    (xa)b = xa·b.
  • We can look at raising two numbers to the same exponent:
    xa ·ya = (xy)a.
  • For any number at all, raising it to the 0 turns it into 1:
    x0 = 1.
  • Raising a number to a negative will "flip" it to its reciprocal:
    x−a =1








  • Because of the above, we see that a denominator is effectively a negative exponent. This means if we have a fraction where the numerator and denominator have the same base, we can subtract the denominator's exponent from the numerator's exponent:


    = xa−b.
  • Raising a number to a fraction is the equivalent of taking a root:
    x[1/2] = √x,                      x[1/3] =


    ,                      x[1/n] =


  • If we want to find the value of raising something to an irrational number, we can find a decimal approximation of the true value by just using many decimals from our irrational number:
    8π    =   83.1415926…    ≈   83.14159.
    The more accurate we need our approximation to be, the more decimals we can use from the irrational number.

Understanding Exponents

Simplify each of the below expressions.
x4·x9               t ·t3 ·t7 ·t8 ·t−5

  • The important rule here is one of the most fundamental rules of exponents: xa ·xb = xa+b. [This rule is true for any variable or expression, we just use x for convenience.]
  • For the first expression, x4 ·x9, the rule says that because they are directly multiplying each other and have the same base, we add the exponents.
    x4 ·x9     =     x4+9     =     x13
  • Things work similarly for the second expression, we just need to repeat the process. We could add the first two together, then the next to the result, then the next, and so on. Alternatively, and more easily, since we're just going to wind up adding all the exponents together, we can add them all together in a single step:
    t ·t3 ·t7 ·t8 ·t−5     =     t1 + 3 + 7 + 8 −5     =     t14
    [If you're not sure where the 1 in the sum comes from, it's because a t on its own can be seen as t1.]

x13               t14
Simplify each of the below expressions.
(x3)2               ( ( ( t2 )5 )[1/4] )2

  • The important rule here is that if you exponentiate something that's already been exponentiated, the exponents multiply: (xa)b = xa·b. [This rule is true for any variable or expression, we just use x for convenience.]
  • For the first expression, (x3)2, the rule says that because we're raising it to the 2 after already raising it to the 3, we multiply the numbers:
    (x3)2     =     x3 ·2     =     x6
  • Things work similarly for the second expression, we just need to repeat the process. We could start with the in-most part of the expression, multiply those exponents, then do that again with the next exponent, then again with the next. Alternatively, and more easily, since we're just going to wind up multiplying them all together, we can multiply them all together in a single step:
    ( ( ( t2 )5 )[1/4] )2     =     t2 ·5 ·[1/4] ·2     =     t5

x6               t5
Simplify each of the below expressions.
x0               47190               ( a + b10 + 28)0

  • The important rule here is that anything raised to the 0 becomes 1: x0 = 1. [This rule is true for any variable or expression, we just use x for convenience.]
  • This is true for anything: if a number or a variable or an expression is raised to the 0, it becomes 1: 47190 = 1.
  • This is true for the last expression as well: ( a + b10 + 28)0 = 1. It might be a little bit confusing because there are multiple things there, but think about it like this: no matter what (a + b10 + 28) comes out to be, it comes out to be a number. When you raise any number to the 0, it becomes 1. So, because the expression is in parentheses, we're raising the whole expression to the 0, so we get 1.

1               1               1
Simplify each of the below expressions.







  • The important rule here is that a negative exponent causes a fraction to "flip". If something isn't already written as a fraction, remember, you can always put it over 1, then flip the top and bottom. From the rules in the lesson, we have x−a = [1/(xa)]. [This rule is true for any variable or expression, we just use x for convenience.]
  • Thus, for the first expression, the negative exponent causes it to "flip" into the bottom of a fraction.
    x−4 = 1

  • Things work similarly for the second expression as well. Because it has a negative exponent, the fraction flips, and then we apply the now-positive exponent to each part:







        =     72

        =     49

  • We continue this process with the third expression, the only difference is that we want to deal with the negative exponents inside the parentheses first. The negative exponent on the u causes it to flip to the bottom of the fraction, while the negative exponent on the v causes it to flip to the top:







    Once that's taken care of, we apply the −3 exponent on the outside of the parentheses to the entire fraction:







        =     u4·3w2·3

        =     u12w6

    [Alternatively, you could also use the rule of exponentiation on exponentiation causing exponents to multiply:




        =     u(−4) (−3)

        =     u12

        =     u12w6

    Either way is acceptable and gets you to the same answer.]

[1/(x4)]               [49/16]               [(u12w6)/(v18)]
Simplify each of the below expressions.
64[1/3]               ( 257 )[1/14]

  • The important rule here is that fractional exponents are connected to roots: x[1/n] = n√{x}. [This rule is true for any variable or expression, we just use x for convenience.]
  • Thus, for the first expression, we get the appropriate root:
    64[1/3]     =    

    Remember, 3√{64} is the cube root of 64: that is, the number that, when raised to the third power, will give us 64. After thinking about it for awhile, we realize that 4·4 ·4 = 64, and so 3√{64} = 4.
  • For the next expression, we might be tempted to begin by figuring out what 257 is. Once we've used a calculator to find that out, we can attempt to take the 14 root of whatever giant number we get. The above would work, but there's a much easier way! Instead, begin by using one of our other rules: exponentiation on exponentiation causes the exponents to multiply.
    ( 257 )[1/14]     =     257 ·[1/14]     =     25[1/2]
    At this point, it's much easier to apply the root rule:
    25[1/2]     =    


        =     5
    [If you're wondering why it's written as √{25} and not 2√{25}, that's because the radical symbol (√{  }) is automatically assumed to be a square root (2√{  }) unless another number is put down to show that it is a different root.]

4              5
Approximate the below to the first two decimal places.

  • It is impossible to write out the precise value of 5π with numbers because it will be an irrational number: one where the decimal expansion continues forever. However, we can get a good approximation by plugging in a decimal number for π, then using a calculator.
  • What number should we use for π? Remember, π is also irrational, so whatever number we use for it, it will also be an approximation. While we could use the approximation of 3.14, the less accurate our π approximation is, the less accurate our end result will be. Therefore, let's use (at least) the first six digits of π: 3.14159. If you want even more accuracy, use even more digits of π.
  • Now that we have a number that we can work with, we can plug it in to our calculator:
    5π≈ 53.14159 = 156.9918748…
    The problem asked for the first two decimal places, so we cut it off to 156.99. [Notice that 5π≠ 156.9918748…. Using a much longer string of digits for π, we can find that
    5π = 156.9925….
    Therefore, while 53.14159 = 156.9918748… is a good approximation, it is not perfect. In fact, no number can be perfect. Because we calculate the value using digits of π, we have to choose some specific number of digits. But since π goes on forever, we can only ever use an approximation of the true value of π, so our final result must also be an approximation. We can get very good approximations, but they can never be absolutely precise. The only way to represent what the number is absolutely precisely is by using what we started with: 5π.]

156.99 [If you got an answer that was close, but not quite the same as the above, you probably need to use more digits of π. Check out the steps for a more detailed discussion.]
Simplify the below expression.
(x2 ·x−10 ·x7 )5

(x80 ·x20 )−[1/5]

  • To do this, we will need to use a combination of the rules we've gone over previously in these questions. These rules can be applied in many different orders, but will all give the same result. The steps below will give a method that is fairly direct and quick, but they are not the only way to find the answer.

  • (x2 ·x−10 ·x7 )5

    (x80 ·x20 )−[1/5]
        =     (x2−10+7)5

    (x80+20 )−[1/5]
        =     (x−1)5

    (x100 )−[1/5]

  • (x−1)5

    (x100 )−[1/5]
        =     x−1 ·5

    x100 ·(− [1/5])
        =     x−5

  • Since both the numerator and denominator each have negative exponents on them, they both "flip" sides:

        =     x20

    At this point, we can just cancel out like usual: x20 means multiplying 20 x's, while x5 means multiplying 5 x's, so the 5 on the bottom will cancel out 5 of those on top:

        =     x15

Simplify the below expression.
173 ·

3−6a ·38a

36a ·176a

  • To do this, we will need to use a combination of the rules we've gone over previously in these questions. These rules can be applied in many different orders, but will all give the same result. The steps below will give a method that is fairly direct and quick, but they are not the only way to find the answer.

  • 173 ·

    3−6a ·38a

    36a ·176a
        =    173 ·

    3−6a + 8a

    36a ·176a
        =     173 ·


    36a ·176a
  • Notice that since 176a does not have the same base as 32a and 36a, it can't really interact with them directly: there is no way to combine them. For now, just set it aside from the fraction (although it must remain inside the radical).
    173 ·


    36a ·176a
        =    173 ·


    · 1

        =    173 ·


    · 1


  • 173 ·


    · 1

        =    173 ·

    3−4a ·17−6a
        =     173 ·(3−4a ·17−6a )[1/2a]

  • 173 ·(3−4a ·17−6a )[1/2a]     =    173 ·(3−4a ·[1/2a] ·17−6a ·[1/2a] )     =    173 ·( 3−2 ·17−3 )

  • 173 ·17−3 ·3−2     =     170 · 1

        =    1 · 1

        =     1


Let f(x) = 17 − x[8/9] and g(x) = x[3/4]. Give (f °g ) (x) and simplify. Then find (f °g ) (27).

  • To find (f °g ) (x), we compose the functions by plugging one into the other. Remember, it's almost always easier to know what to do by writing the above function composition in its equivalent form:

    f °g
    (x) = f
  • Plug g(x) in to f:
        =     f( x[3/4])     =     17 − ( x[3/4] )[8/9]
  • Simplify based on the rules of exponentiation:
    17 − ( x[3/4] )[8/9]     =    17 − x[3/4] ·[8/9]     =     17 − x[2/3]
    Thus, in simplest form, we have (f °g ) (x) = 17 − x[2/3].
  • To find the value (f °g ) (27), just plug x=27 in to what we just found:

    f °g
    (27)     =     17 − (27)[2/3]    
    From there, just simplify based on the rules of exponentiation:
    17 − (27)[2/3]     =     17 −


        =     17 − (3)2     =     17−9     =     8

(f °g ) (x) = 17 − x[2/3],     (f °g ) (27) = 8
Let t be a number such that 64t = [9/4]. What is 8t?

  • Right now, we can't directly figure out what the value of t is. However, we can figure out how 8 connects to 64 in terms of exponentiation. [In a few lessons, once we learn about logarithms, we will be able to solve for t directly. But it would still be easier to do this problem the way shown below.]
  • We can connect 8 and 64 through exponentiation by the following:


    = 8     ⇒     64[1/2] = 8
    Once we see that, we can replace 8 with 64[1/2].
  • Plugging in to our expression 8t, we have
    8t     =     (64[1/2] )t     =     64[1/2] ·t     =     64t ·[1/2]     =     ( 64t )[1/2]
  • We can now replace 64t with what we were given in the problem, then simplify:
    ( 64t )[1/2]     =    



        =      ⎛


        =     3



*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.


Understanding Exponents

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:05
  • Fundamental Idea 1:46
  • Expanding the Idea 2:28
    • Multiplication of the Same Base
    • Exponents acting on Exponents
    • Different Bases with the Same Exponent
  • To the Zero 5:35
    • To the First
    • Fundamental Rule with the Zero Power
  • To the Negative 7:45
    • Any Number to a Negative Power
    • A Fraction to a Negative Power
    • Division with Exponential Terms
  • To the Fraction 11:33
    • Square Root
    • Any Root
  • Summary of Rules 14:38
  • To the Irrational 17:21
  • Example 1 20:34
  • Example 2 23:42
  • Example 3 27:44
  • Example 4 31:44
  • Example 5 33:15

Transcription: Understanding Exponents

Hi--welcome back to Educator.com.0000

Today, we are going to talk about understanding exponents.0002

At this point, we are quite used to using exponents; we have seen them a bunch, and we just did them a whole lot when we were working with polynomials.0005

We know that an exponent is just a shorthand way to express repeated multiplication.0012

For example, if we had 37, that would be a way of just saying 3 times 3 times 3 times 3 times 3 times 3 times 3.0016

That is 3, multiplied by itself 7 times; so the number of times it multiplies is what the exponent is; that is what it represents.0026

x2 is x times x; x3 is x times x times x (three x's); x4 would be four x's multiplied together; etc.0035

So, that is the idea of exponentiation.0044

It is clear how this process works when the exponents are positive integers; it is just multiplied by itself that many times.0047

But if we wanted to expand to any real number--what if we wanted to be able to exponentiate to any real number at all?0052

If we wanted to know things like 30, 37/8, 3-5, or 3√2, how can we deal with that?0058

This lesson will show us how to work with any kind of exponent--any real number.0066

You may have seen rules for this stuff in a previous math class; but we are also going to work through an understanding of why these rules are true.0070

Even if you remember what the rules are for what 30 is, for instance,0077

you might not have a good grasp of why 30 is what it is.0081

And so, this is the lesson that we are actually going to see how we build these rules, where they come from,0085

why we can trust in them, and also how we can make them ourselves, if we ever forget the rules.0089

If we forget the long summary list of all the rules for exponents, we will be able to just go back to our workshop0094

and work it out, even in the middle of an exam; it won't take that long for us to figure out, if we have a sense of how we get these things.0100

All right, at heart, exponentiation is this idea of repeated multiplication; that is the basic, fundamental idea.0107

By definition, for any number x and any positive integer a, xa is equal to x, multiplied by itself a times.0114

This is the key idea behind all of our coming rules for exponents.0123

If we start with this idea, and we just hold it as a fundamental truth and see all the places that it can take us,0126

we will be able to get all of these other cool rules that will explain how it will work for any real number.0132

We take this idea; we say that we believe in this; and then we move forward and try to figure out0137

what else has to be true if this idea right here is always going to work out.0141

Let's see: the first thing we can figure out: with this idea in mind, we can consider what happens when we multiply0147

(that should be "multiply," not "multiple") some xa and xb.0153

By definition, if we have xa times xb, then that means we have xa,0160

so we have x multiplied by itself a times; and xb is x multiplied by itself b times.0165

Now, that means we have some number of x's there and some number of x's there; we have a x's on one side and b x's on the other side.0172

So, how many is it together? Well, let's say, if I had a pile of 5 rocks over here, and I had a pile of 12 rocks over here, then in total, I would have 17 rocks.0179

If I had 3 rocks here and 4 rocks here, then I would have 7 rocks total.0190

So, if we have a rocks and b rocks here, then together that is going to be a + b rocks (or in this case, x's).0196

So, we put them together, and we see that we have a + b many x's showing up.0206

Thus, xa times xb equals xa + b.0211

This is another really fundamental property, and this is what is going to allow us to explore a lot of things in exponentiating with real numbers.0217

We can look at exponentiation acting on top of exponentiation.0226

If we have xa, and then raise that to the b, then that means we have xa multiplied by itself b times.0230

So, this will give us a total of a times b x's; why is that the case?0237

Well, imagine if, in each box, there are 5 rocks; well, if we have 3 of these boxes, then it is 3 times 5.0241

So, if we have a rocks in each box, a x's in each of these boxes (each of these xa's),0251

and we have b many of them, then it must be a times b of them in total.0258

Thus, (xa)b is equal to xab.0263

We just multiply the two exponents in that case.0268

We can also consider what happens if we have two different numbers, each raised to the same exponent.0272

If we have xa times ya, then we have a many x's and a many y's.0276

But we can also shuffle things around: xy is the same thing as yx; we are pretty confident in this.0283

That is one of the cool things about working with the real numbers--that they are commutative.0290

They are allowed to swap their spaces; 2 times 3 is the exact same thing as 3 times 2.0294

So, x times y is the same thing as y times x, which also means that, if we have xx times yy,0299

if we feel like it (because we are allowed to commute--we are allowed to swap the locations in multiplication),0306

then we could switch this to xyxy, which is exactly what we do here.0310

We take these x's and these y's, and we file them together: one x here, one y here, the next x here, and the next y here.0315

So, we will have xyxy; that will show up a total of a times, as well.0323

xa times ya...if we want, we can write it as (xy), that quantity, all raised to the a.0329

Now, let's try to figure out what happens when we raise a number to the 0--our first sort of difficult question.0337

x0 equals what? First, we can write x = x any time we want; just definitionally, that is the idea of equality; you are equal to yourself.0342

So, we can write x = x as x1 = x1.0354

We can put it to an exponent of 1, because that just means that it is itself, just multiplied once--0358

it is just there by itself, because there is only one of them.0363

So, there is nothing wrong with writing x = x as x1 = x1.0365

But not only that--we know that 1 is equal to 1 + 0; 1 + 0 is just 1.0369

So, if we want, we can take this 1 right here, and we will substitute it for 1 + 0; and we have x1 + 0.0376

And then, we will just sort of knock out this one, and leave it as x, just so it is easier to read what is going on.0383

So, we have x1 + 0 = x.0388

This means we have a 0 on the field that we can play with.0392

By our new property, xa + b equals xa times xb; it is our most basic property.0396

Then, we can break this up, and we can take the 1 and separate it from the 0.0402

So, we will have x1 and x0; and we will just write x1 as just x by itself, since that is what it is.0407

We have x(x0) = x; now, as long as x is not equal to 0 (if x is equal to 0, we can't really divide by it easily),0416

we divide by x; we cancel out the x's on both sides, since they show up on both sides.0428

They disappear, and we are left with x0 = 1.0433

There we go: thus, any number, as long as it isn't 0, raised to the 0, becomes 1.0437

So, we take any number at all; we raise it to the 0; it is going to become 1.0445

50 = 1; (-52)0 = 1; 47 million to the 0--you guessed it--equals 1.0449

So, whatever we take, if we raise it to the 0, it becomes 1.0461

Next, what happens when we raise something to a negative number?0466

For ease, let's just figure out what happens with x-1.0469

We begin similarly: x1 = x1; you can't stop me from saying that, just because it is the same thing on both sides.0472

And then, we can also say that 1 = 2 - 1; 2 minus 1 equals 1, so if we want, we can substitute in on this side here; and we have x2 - 1 = x.0479

Now, notice: we have a -1 on the field; so we use that property; we break it apart.0490

And so, by our property, xa + b becomes xa times xb, we have x2 times x-1.0496

So, x2 times x-1 = x; now, we can divide by x2, once again, as long as x is not equal to 0.0507

If we have that, then things get troublesome.0515

But if we divide by x2, the x2 goes from here over to here.0517

And so, we get x-1 = x/x2; and so, x/x2 cancels the x on top0525

and turns this to a 1, and we are left with = 1/x.0534

So, x-1 = 1/x; thus, any number, as long as it isn't 0, raised to a negative, flips to its reciprocal.0538

If we have some number, and we raise it to a negative, we will get that number, flipped into the reciprocal format.0547

5-1 = 1/5...so, what if we want to know what x-a is for any a at all?0555

Well, from the other work that we just did, we know that x-1 to the a is x to the -1 times a, so we can write it in that way, as well.0563

And so, we have 1/xa = 1/xa.0574

A negative in our exponent causes it to flip to its reciprocal format.0581

But it will still keep whatever that original exponent number was, as well.0585

The number will go with it, but the negative is a flip; so negatives flip, but this number that we are exponentiating to will still stay with it.0590

With this idea, we can consider if we had a fraction raised to the -1--not just a number, but a whole fraction.0599

(x/y)-1...well, you can't stop me from separating that into x times 1 over y.0604

And then, we can distribute that -1; remember, xy to the a is equal to xa times ya.0610

So, that -1 will go onto both the x and the 1/y; so we have x-1 times (1/y)-1.0615

x-1 flips to 1/x; (1/y) will flip to y/1, or just y; and so we have y/x.0623

So, negative exponents flip fractions; if we have a negative exponent, we flip whatever it is,0631

whether it is a fraction or a number--we flip to the reciprocal; great.0638

We can also look at if we have powers in the numerator and the denominator with the same base.0642

The base is just the thing that is having that exponentiation happening to it.0647

So, x is our base in almost all of these examples.0652

So, xa over xb...well, we can separate that into xa times (1/x)b.0656

We have xa on top, so we separate that: xa times 1/x, and that whole thing to the b.0662

That is equal to xa times x-b, because 1/x is equal to x-1.0668

So, 1/x to the b is equal to x-b; and now, xa times x-b0675

combines through addition, which, in this case, will become subtraction.0681

So, we have xa - b; thus, the denominator's power subtracts from the numerator's power.0684

That is another thing that we have gotten out of this.0691

Finally, what happens when we raise a number to a fraction?0694

For ease, let's look at just x1/2 first; that will make it easier to understand.0696

Once again, we start from the same place: x1 = x1.0701

And like usual, we want to bring 1/2 to bear; so we notice that 1 = 1/2 + 1/2--it is as simple as that.0704

So now, we can substitute it; we swap this in here, and we have x1/2 + 1/2 = x.0710

Now, notice: we can use our property xa + b = xa times xb,0719

the usual property we have been using, to separate this into x1/2 times x1/2 = x.0725

We already have a name for that--square root.0731

√x times √x = x; that is the idea behind square root.0736

The definition of square root is some number that, when you multiply it by itself, becomes the number you took the square root of.0742

So, √x times √x...√x is just some number that, when you multiply it by itself, becomes x.0749

So, if x1/2 times x1/2 is equal to x, then that must be the same thing as √x,0755

because it does the exact same property--itself times itself becomes x.0763

We already have a name for that: we call that square root.0768

With this property, we see that x1/2 = √x; they are equivalent.0773

We can expand this, by similar logic, to x1/n.0780

x1/n is the same thing as saying that x1/n times itself n times is equal to x,0784

because we have that n many 1/n's is equal to 1.0791

So, if we combine x1/n times x1/n, we will be adding 1/n to itself n times, which is equal to 1.0797

So, x1/n times itself n times is equal to our x, by the same logic that we split up x1/2.0803

By definition, the nth root of a number is something that, when it multiplies by itself n times...we get the original number.0810

The cube root of something, the third root of something, is a number such that, when it multiplies by itself 3 times, we get our original number.0819

The nth root of something is a number such that, when it multiplies by itself n times, we get the original number.0825

Well, look: we are multiplying it by itself n times; we are getting that original number out of it;0831

so it must be that x1/n is equal to the nth root of x.0835

With this idea in mind, we can use any rational number that we want at all.0840

We have xa/b; we can separate that into (xa)1/b.0843

And since we had just had this thing here, 1/n is the same thing as nth root, so we have 1/b becoming b√.0853

We have the b√xa, the bth root of xa.0860

We are just mixing the two properties.0865

The numerator is normal exponentiation, just multiplying by itself, like we normally would expect.0867

And then, the denominator takes a root; it takes that bth root, because it is 1/b.0872

At this point, we have a lot of different rules, and we can see a summary.0879

Our first, foremost, most fundamental rule of all is this idea right here, xa times xb = xa + b.0882

From there, we are able to figure out all of these other rules: (xa)b is equal to multiplying the two together.0891

So, if we have two different exponents raised on one thing, it multiplies them together.0898

If we have xa times ya, we can combine that to just having xy to one a.0904

x0 is always equal to 1, so if you are raising to the 0, you always come out to being 1,0910

as long as we are not dealing with x equal to 0, which we won't address.0917

x-a = 1/xa; we flip with negative exponents.0920

If you have a negative here, then you flip down to the bottom.0925

If we have a fraction with a negative, then the whole fraction flips to its reciprocal; we have y over x to the a.0929

And we also see that xa divided by xb becomes xa - b.0937

Finally, our nth root stuff: x1/n is equal to the nth root of x.0943

xa/b is equal to the bth root of xa,0949

which is the same thing as the bth root of x, to the a, because as opposed to splitting it...0953

we can split this as 1/b to the a, which we would see as b√x, all raised to the a.0958

And that is where we are getting that.0968

So, there are two different ways of looking at it.0969

Sometimes it will be more convenient to have the bth root of xa.0971

Other times, it will be more convenient to have the bth root of x, all raised to the a.0975

It will depend on the specific problem.0979

Remember: I really want you to take away this idea here.0980

If you forget any of these rules, you can figure them out from this fundamental, basic idea: xa times xb equals xa + b.0984

You just have to come up with some creative way to get the thing that you are trying to figure out,0997

whether it is fractions, whether it is 0, whether it is negative numbers--1001

you figure out some creative way to get 0, -1, 1/2, 1/n, something like that, to show up.1006

And then, you look at it, and you say, "Oh, I see--that is what it is!"1011

And so, even if you forget this in the middle of an exam--some place where you can't go and look it up in a book--1014

you can figure this out on your own; it is not that hard.1019

And having worked through it, and understanding how we are getting this, it is that much more likely to stick in your brain.1022

I know it seems like a lot of rules; but once you start using them, and you get used to using them, they will stick in your head.1028

And as long as you remember this one, you can ultimately get back anything that you have forgotten by accident.1034

All right, the final idea: what if we want to raise to an irrational?1039

So far, we have actually only discussed exponentiation using rational numbers.1043

That is the only thing that we have technically dealt with.1048

We have a/b for any a and b, but we haven't dealt with if it can't be expressed as a/b, like √2.1050

So, what if we want to raise something to an irrational number?1057

Let's say we want to look at 3√2; notice: if we want to, we can look at as many places of √2 as we want.1059

We can figure out that √2 is equal to 1.41421356...1067

and it will just keep marching on forever, because it is an irrational number,1072

so its decimal expansion goes on forever, never repeating, always changing, constantly going on forever.1075

But we can figure out what that is.1082

Furthermore, we know how to exponentiate to any rational number.1085

So, we can raise to any decimal, because any decimal is actually something that we can express as a rational.1089

For example, 1.4: if we want to, we can express that as 14 divided by 10.1094

1.414: if we wanted to, we could express that as 1414 divided by 1000.1100

All right, so we can do any of these based on all of the work that we just had.1110

We could do 31.4 as 14/10, 31.414 as 1414/1000...1114

we see that the work we have just done gives us a way to figure these things out.1122

Of course, it would be very difficult for us to do these by hand, but there are methods to do these things.1126

We could do it by hand, but we will leave it to the calculators, since they can do it so much faster.1130

We can use a calculator and get this done so much faster, because they have already been programmed with how these methods work.1134

So, we can take these various things and see: 31.4 becomes 4.6555.1140

31.414 becomes 4.7276; 31.41421 becomes 4.7287; 31.4142135 becomes 4.7288.1146

So, notice: as we use more and more of these decimals, we see the exponentiation, this 3√2, sort of stabilize to a single thing.1160

The 4 always gets used; the 7 always gets used; the 2 always gets used; the 8 always gets used.1169

We see that it is becoming more and more and more stable--that we are seeing more and more of these decimal places show up,1175

and they are not going to change--they are going to stay there forever.1181

So, while we can't get the whole number all at once (it is going to end up being irrational,1184

so it is going to also have a decimal expansion that continues forever, constantly changing), it is stabilizing to something.1189

So, we can get this idea that, while we can't write it down on paper (because it would require an infinite amount of paper), the number does exist.1197

And so, we can get as many decimals as we need for whatever our use is.1205

So, we won't formally define this: but we see that irrational exponents make sense, because we are stabilizing to some number.1209

As long as we use lots and lots of decimals when we calculate it out, 3 to the lots and lots of decimals,1215

from what we were originally trying to use as our irrational number,1221

we will be able to get something that is a very, very close approximation1224

to the exact number that we are trying to strive towards, but won't ever be able to perfectly reach.1228

All right, we are ready for some examples.1233

Evaluate 8/27, all raised to the -2/3.1236

With many of these examples, there are actually going to be multiple ways that we could approach it.1240

So, I will try to show you the various ways that you could go about it.1244

(8/27)-2/3: the first thing I would do is see that we have this negative sign.1247

So, I am going to flip and get rid of that negative: we have (27/8)2/3.1253

That equals...now we have 2 and thirds, so I would put the third into nth roots on both of them.1260

We have the third root, the cube root, of 27, and the cube root of 8.1268

And so, we now no longer have that dividing by 3 to worry about.1272

But we still have the squared, because we didn't get rid of it by putting anything out there.1277

So, the cube root of 27: 3 times 3 times 3 is 27, so we have 3; 2 times 2 times 2 is 8, so we have 2.1281

That is all raised to the 2, once again; so that is 32/22.1292

It gets distributed: 9/4, and there is our answer.1299

But there are also other ways that we could have done this.1304

We could have seen this as (8/27)-2/3, and we could have gone about this as flipping to (27/8)2/3.1306

And we could then put this as [(272)/(82)]1/3; and that is going to be kind of difficult for us to do.1318

So, we could do this with a calculator; and then we could take the cube root of 272 over 82.1327

And that would eventually simplify out to 9/4.1335

But that would be very difficult to do by hand.1338

But notice that we can do the cube root of 27, and we can do the cube root of 8.1340

So, this is probably the much easier way to do this.1343

Furthermore, we could even go about this by taking this as 8/27; we could put this as -2/3 on 8, and then 27-2/3.1346

And then, since they are both negative, they would flip into 272/3 over 82/3.1356

So, we would have 272/3 and 82/3; and then, we could, once again,1362

do either this method here or this method here.1368

At this point, I think pretty clearly that this here is our best bet--the easiest way to do it--1373

where we go through this method, because we see, "8...27...I am going to have to deal with cube roots."1380

8 and 27 are things I can easily take a cube root of, so I am going to do cube root first, then square.1389

And I will also get rid of that negative, as just a first step.1395

You can do these things in many different orders, because the rules all work together.1398

But you will want to get a sense, and as you work on more examples, you will get a sense,1402

of "Oh, the way that will make this problem easiest is for me to go through like this."1407

And so, you will develop an intuition about it.1411

And even if you end up going in a way that is not the easiest, it will still work out.1414

It just might require using a calculator or require a little extra effort.1418

All right, the next one: Simplify (x2/z)-2 times (x2y3z-1)3/y8.1421

All right, the first thing I would do is deal with the negatives, once again.1431

Usually that is easiest; so this will become z/x2, all raised to the now positive 2,1434

times...and let's distribute this 3; the 3 will go onto the x2, the y3, and the z-1.1441

So, it is x to the 2 times 3, because we have exponentiation on exponentiation; y to the 3 times 3, z to the -1 times 3, all divided by y8.1449

We can deal with this squared, and we get z2 over...that 2 also distributes onto the top and to the bottom,1463

so x2 squared is x2 times x2, or x4.1472

Also, it is x to the 2 times 2--another way of looking at it.1477

Times x6y9z-3, all over y8...1480

At this point, we see that we can cancel out the y8, and this becomes y1,1491

which we could also look at as y9 - 8, because we have one on top and one on bottom,1496

which would also become y1; so there are various ways to do this.1502

z2 times x6 times...I will move that over...let's put all of our variables so that they are near their similar ones...1508

We have x6 times y1 (which I will just leave as y), times z2, times z-3, all over x4.1520

And that is all we have on the bottom at this point.1533

So, we see that we have z2, and we see we have z-3; so that will cancel out,1536

and we will get -1, because -3 + 2...z2 times z-3 combine through addition,1539

because they are both just multiplying each other, so we have z2 - 3,1550

which becomes z-1, which is how we get what we have right here.1555

x6 divided by x4 will cancel out all but two of these, which we could also see as x6 - 4, which equals x2.1560

So, we have x2 times y times z-1.1572

And since z-1 is 1/z, we can write this as x2y, all over z.1577

Once again, like our first example, there are other things that we could have done at various points.1585

If we wanted to, we could have broken off here, and we probably could have written this as z2/x4,1589

on our next step, times x6y9, and then z-3/y8.1596

At this point, we see z-3, so we could move that over, and we could do x6/x4,1606

times y9/y8, times z2...1612

and since we had z-3, we could also write that as z3.1616

At this point, we have x6 - 4 times y9 - 8 times z2 - 3.1620

So, we have x2 times y1 times z-1, which also becomes x2y/z.1630

Or, if we wanted to, we could also just say, "We have 9 y's on top and 8 y's on the bottom;1639

so all of them will cancel on the bottom, and one will be left on the top," and similar things with the x's and the z's.1644

So, there are a variety of ways to look at these things, once you get into this.1650

And once again, it is about developing an intuition and just doing it a bunch of times.1653

And also, just be comfortable in the fact that whatever way you choose, as long as you follow the rules, they will all end up working out eventually.1657

The third example: Simplify n√(5n(53n))2n, divided by 56n2.1665

All right, this is a great one to show two different ways to approach this.1674

Let's leave that nth root intact in our first one.1676

nth root of 5n times 53n:1680

well, 5n times 53n, because they are multiplying, will go through addition: 5n + 3n.1683

So, we have 54n, all raised to the 2n, over, still, 56n2.1692

Let's expand that radical a bit, so we see the whole thing.1700

It equals...it still has that nth root, n√(54n(2n),1703

because it was exponentiation on exponentiation, 4n on 2n, over 56n2),1711

equals n√(58) (4 times 2 is 8; n times n is n2) over 56n2.1719

Now, at this point, you might be tempted to cancel out our n2's, but that would be improper,1730

because it is not like canceling 3/3, where we can cancel both of them, because they are dividing.1735

It is different, because it is about how many times they show up.1741

So, we have to use the rule that we have, which is that, when we have a fraction with something on top and something on the bottom,1743

it subtracts if they have the same thing on the bottom.1750

They are both 5's, so they can use this rule.1753

It is still the nth root, so it is going to be 58n2 - 6n2.1756

So, those are common terms, so that becomes...8n2 - 6n2...8 minus 6 is 2, but it still has n2, so it is 52n2.1765

nth root is just the same thing as saying 1/n; so it is 52n2(1/n) = 52n.1776

Great; an alternative way we could have done this, though, is that taking something to the nth root is 1/n, whenever we do it.1791

So, 5n times 53n...once again, that will be 54n, because they add together.1799

And that is all raised to the 2n, over 56n2.1805

Now, it was a radical of the whole thing, so it has to be that the whole thing is raised to the 1/n.1809

Now, we can distribute this, and we can say that that is 54n.1817

Let's leave that as 2n times 1/n, over 56n2 times 1/n,1823

because this 1/n will get applied to the top and to the bottom of our fraction.1833

54n...well, these n's cancel out; this 1/n cancels out the squared and leaves the n, so we have (54n)2/56n.1838

54n(2), because it was exponentiation on exponentiation, and still 56n,1854

equals 58n/56n, equals 58n - 6n, which equals 52n, just as well.1861

Great; so, these are two different ways of doing it, very different approaches--going through the inside,1875

or going from the outside in or inside out--but they both end up giving us the exact same answer.1879

One of the great things about all of these rules is that they all work together.1885

There is no preference of one rule versus the other.1888

So, sometimes it generates various different paths that we could go.1890

But you will develop an intuition; and once again, they all will end up working out.1893

Just make sure you practice these things on your own, and you will develop a sense for how this works.1898

And it will get faster and faster, the more you practice it.1902

All right, another example: f(x) = 7x2/3 - 2; g(x) = x6/5; give f composed with g(x), and simplify.1904

Now, remember: the first thing, when we talked about function composition:1913

f(g(x)) is almost always the way we want to switch to writing this, f acting on g acting on x.1917

What is g(x)? g(x) is x6/5, so it is f(x6/5).1925

So now, it is not about the x; it is about where the box of input goes into our formula for that function.1933

So, f(box) = 7(box)2/3 - 2.1941

In this case, our box is x6/5; we have 7 times box, x6/5;1946

and then, that box raised to the 2/3 - 2, because that is what our whole function said before.1956

So, it is 7 times x6/5; because it is exponentiation on exponentiation, it is multiplication;1962

2/3 - 2; 7 times x; 6/5 times 2/3; we notice that 6 can be broken into 3 times 2, so that knocks out this 3 and this 3.1969

And we are left with 2 times 2 on the top, so that is 4/5, minus 2.1980

So, 7 times x4/5 - 2 is what f composed with g comes out to be.1988

The final example: let x be a number such that 7x equals 3/5; what is 49x?1995

Now, at first, you might see a problem like this, and it is completely confusing, because you have no idea what to do.2002

But notice: we have 7 here; we have 49 here; so there is going to be some sort of clever trick2006

connecting the fact that 49 has something to do with 7.2012

How is 49 related to 7? 49 is equal to 72.2016

So, there is this connection between 49 and 7, so we can use that.2023

We can now apply that, and we can say, "49x...we know that 49 is equal to 72, so we can just substitute that out."2027

So, put 49x in parentheses; it is the same thing as (72)x.2038

There is nothing you can do to just stop substitution like that.2045

So, (72)x--that means 7 to the 2 times x, but we could also write that as 7 to the x times 2,2048

which we could then write as 7x, all raised to the 2, which would be...2057

we know what 7x is--it is 3/5!2062

So, we have 3/5, all raised to the 2, which means we have 9/25, because we square the 3, and we square the 5.2066

3/5 squared means that the square will go onto the 3 and go onto the 5.2078

All right, exponents are pretty cool stuff; they are really, really powerful.2083

It is important to get a good grasp of just working with them, though.2086

The only way that you will be able to get really comfortable with them is doing some practice.2089

So, just make sure that you do some practice with exponentiation, using exponents of various types.2092

But once you get in a bit of practice, you will get used to it; they are skills that stick with you.2097

And as long as you stay with this xa times xb = xa + b,2101

as long as you stay with that idea, you can figure out everything else if you get in a situation where you forget one of the rules.2109

All right, we will see you at Educator.com later--goodbye!2115