WEBVTT
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In this section, we're going to discuss negative exponents. In previous sections, we have
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used and learned rules for whole number exponents, and we're now going to extend our understanding to
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include negative integer exponents. We'll begin with an exploration of x squared divided by x
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to the fifth. Let's consider this expression x squared divided by x to the fifth and simplify
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it using the quotient rule. We've learned previously that when I divide an expression
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by another expression with the same base that I will keep the base and subtract the exponents.
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So this would give me x to the two minus five, or x to the negative three.
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Now considering the same expression x squared divided by x to the fifth,
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I know that I can write this expression as x times x in the numerator and then
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x times itself five times in the denominator to represent x to the fifth.
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Using the fundamental principle of fractions, we can reduce this fraction by factoring out the
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like factors. I know that x divided by itself is just one, so the common factors in this expression
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is an x squared. This will leave us when I simplify this as one over x times x times x, which is x to the third power.
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Well, since we know that x squared over x to the fifth is equal to x to the negative three,
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and it is also equal to one over x cubed. Then that means that x to the negative three must be
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equal to one over x cubed. This leads us to a principle for negative exponents.
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We can say that a to the negative n is equal to one over a to the n for values of a that are not zero and for n's that are integers.
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Let's look at a few examples. We begin with six to the negative two power. The negative exponent
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means that we can rewrite this expression as one divided by six to the positive two power.
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And six to the positive two power, also called six squared, is thirty-six. So this gives us one over
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thirty-six. Keep in mind that a negative exponent does not change the sign of the base in any way.
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In our next example, we have negative three raised to the negative four. We can rewrite this as one divided by
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negative three to the positive four. Rewriting our exponent from a negative exponent to a positive did not change our base.
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Now we need to evaluate negative three to the fourth power, which is negative three times
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itself four times. This will give us a positive eighty-one, resulting here in one over eighty-one.
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And in our last example, we have one-half raised to the negative three power.
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This gives us one divided by one-half to the positive three power.
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Now recalling our rules for exponents of quotients, this three can be applied to both
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the one and the two in the fraction of one-half, giving us one divided by one cubed over two cubed.
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We know that dividing by a fraction is the same as multiplying by a reciprocal,
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so we can rewrite this again as one times the reciprocal of one cubed over two cubed,
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which would give us two cubed over one cubed, or eight divided by one, which is eight.
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All of the previously stated rules for exponents apply also for negative exponents. Let's look at
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these next few problems which will combine all of the exponent rules that we've learned so far.
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We want to simplify the expression by rewriting using only positive exponents.
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We begin with four a to the negative nine. Let's begin by noticing what is it that's
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being raised to the exponent, negative nine. It is only the a. So the four will remain as it is,
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and a to the negative nine can be written as one over a to the ninth.
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So we see that we have four times one over a to the ninth, which is the same as four over
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a to the ninth. We've rewritten the problem with only a positive exponent.
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In the next problem, we have three divided by b to the negative five. Again, notice that b is being raised to the exponent,
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negative five. So the three will remain in the numerator. b to the negative five can be written as one
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over b to the fifth. And recall that dividing by a fraction is the same as multiplying by
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its reciprocal, so this will be the same as three times the reciprocal of one over b to the fifth,
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which is just b to the fifth, leaving us with three b to the fifth power.
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In our third example, we have x to the negative eight divided by x squared.
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Let's use our quotient rule to simplify to begin with. We have the same base
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on these expressions, and we're dividing them, so we subtract the exponents.
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We keep the base x, and we have negative eight minus two as our new exponent, giving us x
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to the negative ten. Well, our directions tell us to rewrite using positive exponents,
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so we can rewrite again. x to the negative ten is the same as one over x to the positive ten.
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Our next example is fifteen p to the sixth q divided by three p to the negative one
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q to the fourth. Let's begin by separating these factors by their bases, by their like bases.
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I'll start with my integers however and combine those into one expression. Fifteen
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divided by three can be grouped, and then p to the six divided by p to the negative one,
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and q to the first divided by q to the fourth. Notice I added a one as the exponent on this q.
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There is an implied exponent of one whenever we don't see an exponent.
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Now let us simplify each of these factors. First, fifteen divided by three is five.
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p to the sixth divided by p to the negative one is p to the six minus
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a negative one. And q to the first divided by q to the fourth is q to the one minus four.
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This gives us five p to the six plus one, or seventh, power q to the negative three power.
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And in our final step because we want to rewrite this expression with only positive exponents,
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we notice that we have a q to the negative three which is the same as one over q to
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the third power. So five p to the seventh will remain in the numerator of our answer,
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and q to the third power will be in the denominator of our answer.
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In our next example, we have x squared y to the fourth raised to the negative two power.
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Here we have a product being raised to a power. So we can apply this power of negative two to each of these factors.
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This gives us x squared to the negative two times y to the fourth to the negative two.
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And now we have a power raised to a power, and we know to multiply these exponents.
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So this gives me x to the negative four y to the negative eight. And now as I simplify these expressions, or this expression,
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each of these factors are raised to a negative power which means x to the negative four will become a one
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over x to the fourth, putting that factor in the denominator. And y to the negative eight is the same as one over y to the eighth,
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meaning that y to the eighth will be in the denominator. And that leaves us with a one in the numerator.
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In our last example, we have two a b to the negative three times negative four
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a to the negative three b squared. So we're going to multiply these two factors together.
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We know that by the Commutative and Associative Property of Multiplication that we can rearrange
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these factors and regroup them. So let's do that. We will multiply two times negative four
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as one grouping and a times a to the negative three as another grouping and b to the negative three times b squared
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as our third grouping. Now let's apply our rules of exponents and multiplication. Two times negative four
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is negative eight. a, remember to the first power is implied, times a to the negative three will give us a to the
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one plus a negative three in the exponent. And then b to the negative three times b squared will give us
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b to the negative three plus two. This gives us negative eight a to the negative two b to the negative one.
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And finally, to clear our negative exponents, we recall that a to the negative two is the same as
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one over a squared, and b to the negative one is one over b. So these factors will
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move to the denominator when we write our final answer, negative eight divided by a squared b.