Lesson Video: Evaluating Expressions: Exponents | Nagwa Lesson Video: Evaluating Expressions: Exponents | Nagwa

# Lesson Video: Evaluating Expressions: Exponents Mathematics

In this video, we will learn how to evaluate algebraic expressions with variable exponents.

14:07

### Video Transcript

In this video, we will practice evaluating algebraic expressions with variable exponents. We might first want to remind ourselves what an expression is. Expressions are numbers, symbols, and operators grouped together to show the value of something. For example, if we have the equation three plus five to the π₯ power equals 128, three plus five to the π₯ power is an expression. In this expression, we have five to the π₯ power. We have an unknown value as the power. In this video, we will specifically be looking at algebraic expressions where solving them will include solving for a variable as the power or the index, exponents where the power is unknown.

But before we get to that, we need to do a quick review of the exponent rules and working with exponents. When weβre multiplying exponents, π₯ to the π power times π₯ to the π power equals π₯ to the π plus π power. When weβre dividing exponents, π₯ to the π power divided by π₯ to the π power equals π₯ to the π minus π power. Taking a power of a power, π₯ to the π power to the π power equals π₯ to the π times π power. And then we have the power of a product, which would be something like π₯ times π¦ to the π power, which would yield π₯ to the π power times π¦ to the π power.

And then we have the power of a fraction. If we have π₯ divided by π¦ to the π power, weβll have π₯ to the π power divided by π¦ to the π power. Next, we have a zero exponent. π₯ to the zero power equals one. Anything taken to the zero power equals one. And for our negative exponents, π₯ to the negative π power is equal to one over π₯ to the π power. Finally, if we have a fractional exponent, sometimes called rational powers, we would have π₯ to the one over π power. And that is equal to the πth root of π₯. Using these rules, weβll be able to simplify and solve for variable exponents. Letβs consider some examples.

Evaluate three plus five to the π₯ power if π₯ equals two.

In this case, we have the expression three plus five to the π₯ power. In order to evaluate this expression, we substitute the number two in for π₯ so that we have three plus five squared. Five squared is five times five, which is 25. We add three to 25 to get 28, which tells us that three plus five to the π₯ power is equal to 28 when π₯ equals two.

In our next example, weβll consider a case where weβre dealing with two different variables.

Evaluate π₯ to the π§ power minus π§ to the π₯ power, where π₯ equals four and π§ equals three.

Weβre given the algebraic expression π₯ to the π§ power minus π§ to the π₯ power. Itβs possible at this point that you might think π₯ to the π§ power minus π§ to the π₯ power must be zero. However, this is not true. π₯ to the π§ power does not equal π§ to the π₯ power. They are not additive inverses of each other. And we can confirm this by plugging in the values that we were given for π₯ and π§. Since we know that π₯ equals four and π§ equals three, our first term will be four cubed and our second term will be three to the fourth power. Four cubed is equal to 64, and three to the fourth power equals 81. This expression would then be 64 minus 81, which equals negative 17. And we see here that four cubed, π₯ to the π§ power, does not have the same value as three to the fourth power, π§ to the π₯ power.

In our third example, weβll look at two fractional values being multiplied together when they have different variable exponents.

Calculate the numerical value of five-fourths to the π¦ power times five-fourths to the π₯ power when π₯ is equal to negative seven and π¦ equals four.

We have the expression five-fourths to the π¦ power times five-fourths to the π₯ power. When we look at this expression, we should notice that both of our terms have five over four as their base. And theyβre being multiplied together. This should make us think of our exponent rule π₯ to the π power times π₯ to the π power is equal to π₯ to the π plus π power. This means we can rewrite our expression with the base five over four and a new exponents of π¦ plus π₯. This step simplifies our expression into one term. And from there, weβll substitute negative seven in for π₯ and four in for π¦. We would then have five-fourths to the four plus negative seven power, which we can simplify to the four minus seven power.

But four minus seven is negative three, which leaves us with five-fourths to the negative three power. We can remember the power of a fraction rule, which tells us that π₯ over π¦ to the π power is equal to π₯ to the π power over π¦ to the π power. When we distribute this negative three power, we get five to the negative three power over four to the negative three power. And we also know that π₯ to the negative π power is equal to one over π₯ to the π power. A corollary of this tells us one over π₯ to the negative π power is equal to π₯ to the π power, which means that if we have a negative exponent in the denominator, we can move it to the numerator and make it positive. Five to the negative three power over four to the negative three power can be rewritten as four cubed over five cubed, which equals 64 over 125.

In our fourth example, we wonβt be told exactly what our exponent values are. Instead, weβll have to use given information to calculate the values of variables in the exponent.

Given that two to the π₯ power equals eight to the π¦ power which equals 512, determine the value of π₯ minus π¦.

Before we can solve for π₯ minus π¦, weβll have to determine what π₯ and π¦ are equal to. This statement tells us that two to the π₯ power equals 512 and eight to the π¦ power also equals 512. We could break this up into two separate statements, like this. And thinking about solving each of these, if 512 is equal to two to some power and eight to some power, we know that 512 will be divisible by both two and eight. Since eight is the larger of the two factors, one thing I would do is to try and divide 512 by eight. When I do that, I see that eight times 64 equals 512. And I already know that 64 equals eight times eight.

This shows me that 512 is equal to eight cubed. In our statement, where we see 512, we can substitute eight cubed since 512 equals eight cubed. And then we get a new statement. It says eight to the π¦ power equals eight cubed. And based on this information, we can say that π¦ must be equal to three. Now, weβve already found out that 512 equals eight cubed. But when we have the statement two π₯ equals eight cubed, weβre still not in a position to solve for π₯. But we could be thinking that eight is divisible by two. We know that two times four equals eight. And then two times two equals four. And that means we could rewrite the value eight as two cubed, which means 512 equals two cubed cubed.

Taking what we know about power to a power, π₯ to the π power to the π power equals π₯ to the π times π power. Three times three is nine, which means 512 is equal to two to the ninth power. And our new statement says two to the π₯ power equals two to the ninth power. Therefore, π₯ must be equal to nine. Weβre now ready to solve the expression π₯ minus π¦. We know π₯ equals nine and π¦ equals three. So we can say that π₯ minus π¦ equals six; nine minus three equals six.

In our final example, again, weβll have to solve for the value of a missing exponent. And then weβll use that information to determine the value of an expression.

Given that two to the π₯ power equals four, determine the value of eight to the π₯ minus one power times two to the two π₯ plus six power times one-half to the two π₯ power.

Weβve been given a statement and then an expression to solve for. In order to solve the expression, weβll need to take the statement two to the π₯ power equals four. Using this information, we can solve for π₯. We want to rewrite four with a base of two. Since we know that four equals two squared, we can substitute two squared into this statement in place of four, which allows us to see that two to the π₯ power equals two squared. Therefore, π₯ must be equal to two. Once we know that, weβre ready to consider our expression, eight to the π₯ minus one power times two to the two π₯ plus six power times one-half to the two π₯ power.

In this expression, everywhere we see the variable π₯, weβll substitute the value two. Eight to the two minus one power would be eight to the first power, which is eight. Two to the two times two plus six power β β two times two is four plus six is 10. Weβll have two to the tenth power. And one-half to the two times two power is one-half to the fourth power. Now, at this point, it is completely possible to plug this into the calculator or try and solve for all of these values.

However, there is a way we can simplify a bit first if we recognize that all three of these factors can be rewritten with a base of two. Just like we rewrote four as two squared, we could rewrite eight as two cubed. Two times two times two equals eight. Two to the tenth power already has a base of two. And now, you might be wondering, βbut how will we write one-half as a base of two?β Remember that π₯ to the negative π power equals one over π₯ to the π power. This means we could think of one over two as one over two to the first power. And we can rewrite this as two to the negative first power.

Instead of one-half to the fourth power, weβll have two to the negative first power to the fourth power. And because we know a power of a power rule, π₯ to the π power to the π power equals π₯ to the π times π power, two to the negative one power to the fourth power can be rewritten as two to the negative four power. And one final step for simplification, π₯ to the π power times π₯ to the π power equals π₯ to the π plus π power. Since all of our factors have a base of two and theyβre being multiplied together, we take the base of two and then we add all the exponents together. It will be two to the three plus 10 plus negative four power. Three plus 10 is 13, minus four is nine, two to the ninth power. And two to the ninth power equals 512.

Before we finish, letβs remember a few key points for evaluating expressions with variable exponents. When evaluating algebraic expressions with variable exponents, use the rules of exponents to simplify as much as possible. In addition to that, where possible, rewrite values so that they have the same base. Then use power of a power rule to further evaluate.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions