Lesson Video: Simplifying Expressions: Rules of Exponents | Nagwa Lesson Video: Simplifying Expressions: Rules of Exponents | Nagwa

# Lesson Video: Simplifying Expressions: Rules of Exponents Mathematics

In this video, we will learn how to simplify algebraic expressions using the rules of exponents.

17:20

### Video Transcript

In this video, weβre going to simplify expressions using rules of exponents.

The first question that comes up is, what is simplifying expressions? By simplifying, we mean writing an expression in the most compact or efficient manner without changing the value of the expression. For this process, we might do things like removing parentheses or brackets by multiplying factors, combining like terms, and, as weβll look at today, simplifying exponents. And why would we use this kind of process?

Simplifying expressions makes them easier to read and understand. It also reduces possible errors in calculating, as after an expression is simplified, there are less calculations that will need to occur. Letβs say, we have an exponent in the form π times π₯ to the π power. In this case, the constant in front of the π₯ will be the coefficient. Whatever is being taken to a certain power will be the base of the exponent. The superscript in the upper right-hand corner of the base is called the power or the index. And when all of this together is inside an expression, it will be one of the terms in the expression.

Now, weβre ready to consider some rules about how we operate with exponents. Hereβs the first rule: π₯ to the π power times π₯ to the π power equals π₯ to the π plus π power. We might see something like this: π₯ squared times π₯ cubed. Weβre able to simplify that by writing π₯ to the two plus three power, π₯ to the fifth power. Another rule to consider is π₯ to the π power over π₯ to the π power which is equal to π₯ to the π minus π power. If we had something like π₯ to the sixth power over π₯ to the fourth power, we could simplify that by saying π₯ to the six minus four power, or simply π₯ squared.

The next rule we call a power of a power. π₯ to the π power to the π power is equal to π₯ to the π times π power. For example, π₯ to the seventh power to the third power equals π₯ to the seven times three power, or simply π₯ to the 21st power. Before we move on, with this rule, we can also note that π₯ cubed to the seventh power will also be equal to π₯ to the three times seven power, which is π₯ to the 21st power. And that means π₯ to the seventh power cubed is equal to π₯ cubed to the seventh power.

Next, we have π₯ times π¦ to the π power will be equal to π₯ to the π power times π¦ to the π power. π₯π¦ cubed is equal to π₯ cubed π¦ cubed. We should also say that this rule can be combined with a power to a power rule. If we have π₯ squared times π¦ to the fourth to the third power, weβll have π₯ to the two times three power times π¦ to the four times three power, π₯ to the sixth power and π¦ to the 12th power.

The same principle holds true with division. π₯ divided by π¦ to the π power is equal to π₯ to the π power over π¦ to the π power. π₯ over π¦ cubed equals π₯ cubed over π¦ cubed. And then, π₯ to the zero power is one and π₯ to the first power is π₯. Using these rules, weβre ready to jump in and simplify some expressions.

Simplify π₯ to the 12th power times π₯ to the eighth power over π₯ times π₯ to the 11th power, where π₯ cannot be equal to zero.

This is our expression. Looking at this expression, we should recognize that when we have the same base π₯ to the π power times π₯ to the π power will be equal to π₯ to the π plus π power. In our numerator, that means we can add the exponents 12 and eight to get 20. In our denominator, we see an exponent of 11. And we also know that π₯ is equal to π₯ to the first power. And that means we can rewrite π₯ to the first power times π₯ to the 11th power as π₯ to the 12th power.

And from there we think, will π₯ to the π power over π₯ to the π power is equal to π₯ to the π minus π power. And that means we can take the power in the numerator, 20, and subtract the power in the denominator, 12. When we simplify that, we get π₯ to the eighth power. Since we do not know what π₯ is, we canβt calculate the final answer, but π₯ to the eighth power is in its simplest form.

Before we move on, I want to point out one other way you couldβve solved this problem. In the numerator, we have π₯ to the 12th power. And in the denominator, we have π₯ to the first power times π₯ to the 11th power. If you recognize that one plus 11 equals 12, then you wouldβve seen that the numerator and the denominator both had a factor of π₯ to the 12th power which cancel each other out. Either way, you end up with the simplest form π₯ to the eighth power.

In our next example, weβre given an equation, and weβre going to use that equation to help us simplify and then calculate another expression.

Given five to the π₯ power equals two, determine the value of 25 to the π₯ power.

We have the equation five to the π₯ power equals two and the expression 25 to the π₯ power. We want to know what the value of that expression is. At first glance, we notice that five and 25 are not the same base. And that makes it difficult to do any kind of simplification. But when we think about that 25 in relation to five, we could say that 25 equals five squared. In our expression, in place of 25, we can write five squared so that we have five squared to the π₯ power. Which reminds us that π₯ to the π power to the π power equals π₯ to the π times π power, which would be five to the two times π₯ power.

But at this point, we still may be puzzled because we only know what five to the π₯ power is; we donβt know what five to the two π₯ power is. However, we can do some regrouping. We know that five to the two π₯ power will also be equal to five to the π₯ power squared. What weβre saying is, five squared to the π₯ power will have the same value as five to the π₯ power squared. This is true because theyβre both equal to five to the two π₯ power.

At this point, we know what five to the π₯ power is equal to. Since five to the π₯ power equals two, we can calculate five to the π₯ power squared as two squared, which equals four. Given that five to the π₯ power equals two, the value of 25 to the π₯ power is four.

In our next example, weβll need to simplify an expression to help us calculate its numerical value.

Given that π§ equals negative one-fourth, find the numerical value of negative eight π§ to the fourth power times π§ squared all over 16π§ cubed.

Starting with this expression, we want to get it in its simplest form before we try and calculate its numerical value. In simplifying, we want to get rid of these parentheses. We remember that π₯ times π¦ to the π power equals π₯ to the π power π¦ to the π power so that we have negative eight to the fourth power times π§ to the fourth power times π§ squared. When we bring over the denominator 16π§ cubed, we can do some canceling between the numerator and the denominator. π₯ to the π power over π₯ two to the π power equals π₯ to the π minus π power.

Since we have π§ to the fourth power in the numerator and π§ cubed in the denominator, the π§ cubed in the denominator cancels out and the π§ to the fourth power in the numerator becomes π§ to the first power. Four minus three is one. There are no more π§-variables in the denominator, but in the numerator we have π§ to the first power times π§ squared. We can simplify that by writing it as π§ cubed; one plus two is three. Now, we have negative eight to the fourth power times π§ cubed over 16.

We can no longer simplify that π§-variable. But it is worth considering if we can simplify the negative eight to the fourth power over 16. I know that negative eight could have a factor of negative four and two. If I write negative eight in this way, we would then have negative four to the fourth power times two to the fourth power times π§ cubed over 16, which is helpful to ask because two to the fourth power is 16. So, the factor of 16 in the numerator and the factor of 16 in the denominator cancel out.

We now have negative four to the fourth power times π§ cubed. And weβre ready to plug in negative one-fourth for π§. Negative four to the fourth power times negative one-fourth to the third power. If we have π₯ over π¦ to the π power, thatβs equal to π₯ to the π power over π¦ to the π power. If we break up this fraction, weβll have negative four to the fourth power times negative one cubed over four cubed. I wanna break up this negative four to the fourth power one final time so that we have negative one to the fourth power times four to the fourth power.

Negative one to the fourth power equals one. And since we have four cubed in the denominator and four to the fourth power in the numerator, we can say that weβll have four to the first power in the numerator. We subtract those exponents. Four minus three is one. Four to the first power equals four. And the final operation we need to do is negative one cubed. Thatβs negative one times negative one, which is positive one, times negative one, which will be negative. So, our final answer here, the numerical value for this expression is negative four.

Again, in the next example, weβll have to do a significant amount of simplifying before we can answer the question.

Which of these equations is satisfied by π₯ equals two-thirds cubed squared and π¦ equals three times three-halves to the fifth power minus three-halves to the sixth power? (A) π₯ equals π¦, (B) π₯ equals π¦ squared, (C) π₯ equals one over π¦, or (D) π₯ equals one over π¦ squared.

Letβs start with π₯. π₯ equals two-thirds cubed squared. Using the power to a power rule, we could rewrite this as two-thirds to the sixth power. And from there using another power rule, we can simplify further to say two to the sixth power, three to the sixth power. Now that this value is in its simplest form, we donβt need to calculate yet. We can just leave it as two to the sixth power over three to the sixth power.

π¦ equals three times three-halves to the fifth power minus three-halves to the sixth power. And at first, itβs not immediately clear what can be done here. But one thing we can do is break up this three-halves to the sixth power, using the principle that π₯ to the π power times π₯ to the π power equals π₯ to the π plus π power. We know that six equals five plus one. And that means three-halves to the first power times three-halves to the fifth power is equal to three-halves to the sixth power. And then we have three times three-halves to the fifth power minus three-halves to the first power times three-halves to the fifth power.

And looking at these fractions, again, it might not seem clear what we should do. But imagine, if we replace this three-halves with the variable π₯. It becomes a little bit easier to see that both of these terms have a factor of three-halves to the fifth power. If you undistributed π₯ to the fifth power, youβd have π₯ to the fifth power times three minus π₯ to the first power. We want to follow this principle using three-halves to the fifth power, which means weβll have three-halves to the fifth power times three minus three-halves to the first power.

Now, three-halves to the first power is just three-halves. We wanna do the subtraction within the brackets. And that means weβll need to rewrite three with the denominator of two. Three is equal to six over two. Six-halves minus three-halves equals three-halves. And now, we have three over two to the fifth power times three over two, which is three over two to the first power. And weβre able to use this same principle to make this three-halves to the sixth power, which weβll rewrite as three to the sixth power over two to the sixth power.

π¦ equals three to the sixth power over two to the sixth power and π₯ equals two to the sixth power over three to the sixth power. What weβre seeing is that π₯ and π¦ are the inverse of one another. When weβre working with fractions, the inverse switches the numerator and the denominator. And if π₯ and π¦ are the inverse of one another, then π₯ equals one over π¦.

If you wanted to check that this was true, we have a statement that says two to the sixth power over three to the sixth power is equal to one over three to the sixth power over two to the sixth power. Thatβs saying one divided by three to the sixth power over two to the sixth power. We could rewrite that as one divided by three to the sixth power over two to the sixth power. And we know dividing by a fraction is the same as multiplying its reciprocal. So that would be one times two to the sixth power over three to the sixth power, which is true. For these equations, π₯ is equal to one over π¦.

In our final example, weβll be working backwards from something thatβs already simplified to find its previous form.

Complete the following: 14π₯ cubed π¦ to the fifth power over something is equal to two π₯ over six π¦.

When we look at this equation, whatβs on the right is the simplified form of the fraction on the left. Weβll need to work backwards to find out what the denominator of the first fraction was. Since weβre working backwards, Iβm gonna flip the order so that we have two π₯ over six π¦ is equal to 14π₯ cubed π¦ to the fifth power. Even though we have variables and their coefficients, the principles of keeping fractions proportional still apply. And that means in the numerator, if we went from two to 14, we would have to multiply by seven. But if we multiplied by seven in our numerator, weβd have to multiply by seven in our denominator to keep the values equal.

Six times seven is 42. Going back to our numerator, we have an π₯ going to an π₯ cubed. Here, that means weβve multiplied by π₯ squared. And if we multiply by π₯ squared in the numerator, we have to multiply by π₯ squared in the denominator. Since the denominator doesnβt have an π₯ squared, itβs one times π₯ squared, which is π₯ squared. And then in our numerator, we also have π¦ to the fifth. We donβt have that on the left. So thatβs just going to be multiplying by π¦ to the fifth. We do have a π¦ in the denominator. So, weβre multiplying the denominator by π¦ to the fifth, which is π¦ times π¦ to the fifth, which is π¦ to the sixth power.

Because we donβt generally work in that direction, itβs probably a good idea to check the result. Both 14 and 42 are divisible by seven. That gives two-sixths. The π₯ squared in the denominator is canceled out and leaves us with π₯ to the first power in the numerator. When we look at our π¦-variables, the π¦ to the fifth power in the numerator cancels out. And weβre left with π¦ to the first power in the denominator, which is two π₯ over six π¦ and confirms 42π₯ squared π¦ to the sixth power as the missing denominator.

When it comes to the key points here, simplifying expressions allows us to write an expression in the most compact and efficient manner without changing the value of the expression. And when those expressions have exponents, we can use these basic rules to help us simplify.

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