Textbook

*Exponents*, or powers, help us represent multiplication of the same factor. For example, if we want to express one hundred 10’s multiplied together, we don’t have to write out the number 10 one hundred times! We can simply write it as $10_{100}$. Because it prohibits calculator usage, the CLT will generally keep the powers small so not too much mental or manual calculation is required.

Which of the following is equivalent to $(c_{4})(d_{−1})(c_{−2})(d_{3}) $?

A. $c_{2}d_{2} $

B. $c_{6}d_{2} $

C. $c_{2}d_{4} $

D. $c_{6}d_{4} $

Let’s begin with the concept of the exponent, or power. The purpose of an exponent is to tell us how many *factors* are present. If we know that factors are the terms involved in multiplying, we could say more informally that the power tells us “how many of the base there are.” If we see $2_{4}$, we know that means “four twos multiplied together.” If we see $x_{7}$, that would be seven instances of $x$ as a factor - seven x’s multiplied. That’s why, when we apply the quotient rule (see below), we understand that $z_{2}z_{6} $ will be $z_{4}$, not $z_{3}$, because there are six $z$’s on top and two $z$’s on bottom. Two pairs of $z$’s cancel each other, leaving four “$z$” factors remaining.

CLT exponents questions can combine different rules of exponents, as we see in this example. In this lesson, we have provided a summary of the key rules for your review (flashcards are highly recommended!), but we will discuss the rules relevant to this question here.

The quotient rule of exponents tells us that when dividing identical bases, we *subtract* the powers. So this question calls on us to simplify $c_{4}c_{−2} $ to $c_{−2−4}$, which equals $c_{−6}$. For the other base, meanwhile, we subtract the exponent in the denominator from the exponent in the numerator; $3−(−1)$ is $4$, so our simplified form here is $d_{4}$. Remembering that exponents are a way of expressing multiplication, and that terms placed immediately next to each other are multiplied ($xyz$, for example, means $x$ times $y$ times $z$), we multiply our simplified terms, placing them together as $c_{−6}d_{4}$.

Are we done? You may think so, but a principle used in working with exponents is that an expression with negative exponents is not considered to be simplified. The CLT will usually (though not always) follow this principle. Further, we can see from our UnCLES-based review of the answer choices that none of them have negative exponents, so we can assume that we must simplify further. To do so, we need to understand the rule for negative exponents: they create *reciprocals*. In other words, the base and the positive power will move from numerator to denominator (or vice versa) while the negative power disappears once it has “done its job.” This means that $c_{−6}$ is the same as $c_{6}1 $; creating a fraction to accommodate this new denominator, we now have $c_{6}d_{4} $. The answer is **D**.

Another way of looking at the problem focuses more on the number of factors involved in each power. Starting with the variable $c$, we could say that there are already four “$c$’s” in the denominator and two more that “want” to join them there because of the negative in the numerator. That’s a total of 6 “$c$” factors in the denominator. Meanwhile, there are already three “$d$’s” in the numerator and one more in the denominator where the negative exponents “want” to send them up to the numerator. That’s a total of four “$d$” factors in the numerator. So $c_{6}d_{4} $.

Exponents can be involved in *symbol* problems; since these types of problems are fairly rare on the CLT, we will cover them here rather than in a separate lesson.

In a CLT symbol problem, the test maker uses a random symbol (one without a common mathematical meaning) to direct the student to perform one or more operations. Symbol problems are a little like functions in that you get the “rules of the game” and you simply have to apply the rules. Applying the rules always happens in the order given; the first number or variable given goes first and the second one goes second.

A simple symbol problem would tell you something like $x$ ⇩ $y$ = $x_{2}×y_{3}$ and then ask you to calculate 3 ⇩ 2. Since 3 corresponds to $x$ (both given first), we square it to make 9. Since 2 corresponds to $y$ (both given second), we cube it to make 8. The problem instructs us to multiply the two results, so the answer is 72.

Another variation in exponent problems is a word problem using a power. Although not common, when it does occur, a word problem of this type might present a quantity that is tripled a certain number of times, calling on you to recognize that tripling a number $x$ times is the same thing as multiplying that number by $3_{x}$. In other words, tripling repeatedly is the same thing as 3 $×$ 3 $×$ 3 … Just don’t forget that the initial number tripled might not *itself* be equal to 3. If we start with 4 and triple it three times, we end up with $4×3_{3}=4×27=108$.

Here are all the exponent rules you’ll need to know on the CLT. Although these rules are usually presented with multiple variables involved, we will make the rules more concrete by using $x$ (and sometimes $y$) only as the base and using numbers as the powers.

IMPORTANT: the first two rules (product rule and quotient rule) only work when the bases are the *same*.

- $x_{2}×x_{3}=x_{5}$ (product rule)
- $x_{2}x_{6} =x_{4}$ (quotient rule)
- $(x_{3})_{3}=x_{9}$ (product of powers or what we like to call “raised up high, multiply”)
- $x_{3}×y_{3}=(xy)_{3}$ (in other words, the power remains the same and the bases are multiplied)
- Similar to immediately above, $x_{3}÷y_{3}=(x/y)_{3}$ (the power remains the same and the bases are divided)
- $x_{0}=1$
- $x_{−2}=x_{2}1 $ (and, conversely, $x_{−2}1 =x_{2}$)
- The cube root of $x_{4}$ is $x_{(}34 )$ (the number in the fractional exponent remains a power; the denominator names the number of the root)

What is the value of $2_{3}4_{3} $?

A. 1

B. 2

C. 4

D. 8

(spoiler)

The answer is **D**. If you know that $4_{3}$ is 64 and $2_{3}$ is 8, you can simply divide 64 by 8 to get the answer. If those calculations are not immediate for you, it might be quicker to apply the rule for dividing different bases with an identical exponent: retain the exponent and divide the bases. This yields $2_{3}$, which is 8.

Simplify the expression $a_{5}(a_{4})_{3} $.

A. $a_{2}$

B. $a_{5}$

C. $a_{7}$

D. $a_{12}$

(spoiler)

The answer is **C**. Two different rules apply here. The product of powers rule (“raised up high, multiply”) tells us we *multiply* the powers in the numerator (careful, don’t add!), making $a_{1}2$. The quotient rule for dividing $a_{1}2$ by $a_{5}$ tells us to subtract the powers, leaving $a_{7}$.

If the operation ⟴ is defined such that $m$ ⟴ $n$ = $n_{2}m_{3}n $, which of the following is equivalent to 6 ⟴ 4?

A. 216

B. 54

C. 27

D. 12

(spoiler)

The answer is **B**. With symbol problems, we follow instructions in order. The rules tell us to take the first element (before the symbol), raise it to the third power, and multiply it by the second element … then divide that result by the second element squared. The calculation could get a little complicated here, so look out for ways to streamline the process.

The first thing to notice is that the definition *itself* can be simplified by canceling $n$ on the top and bottom, leaving simply $m_{3}/n$. Second, instead of finding 6 to the third power first, keep in mind that nothing in the order of operations (PEMDAS) means we have to finish multiplying before we divide. We could multiply 6 $×$ 6 to make 36 and then divide by 4 to reduce to 9, only then multiplying by 6 one more time to make 54.

Which of the following is the cube root of the expression $(−3)_{3}×(−2)_{6}×(−3)_{−3}×(−2)_{−3}$?

A. 1

B. -2

C. -8

D. 64

(spoiler)

The answer is **B**. First of all, use the UnCLES method well, noting that we’re not only simplifying the expression but also finding its *cube root*. You can count on there being a trap answer that will catch you if forget to take the cube root!

To simplify, we’ll apply the product rule twice, first with the -3 base and then with the -2 base. Adding the powers of -3 actually gives us zero. Do you remember that anything to the zero power is 1? That means those two terms essentially cancel out, since multiplying a result times 1 doesn’t change anything. Using the product rule on the -2 bases leaves $(−2)_{3}$. We could calculate that, but if we remember we want the cube root, why not just take the cube root here, thereby canceling the exponent? The answer is -2.

If $3x+2=a$, which of the following is equivalent to $2_{2−a}(2_{−3x}) $?

A. 1

B. 2

C. 4

D. 8

(spoiler)

The answer is **A**. At first glance, the equation presented at the beginning does not appear relevant to the expression we’re supposed to simplify. Using UnCLES and considering the answers, we know that somehow we should be able to get a number answer. So, confident of that fact, we proceed to use our tools, the exponent rules available to use. This question requires the quotient rule, since two identical bases are being divided.

But first: should we plug in $3x+2$ for $a$? It can be done now or later, but in general we should take care of substitution as soon as possible. Being carefully to plug in the expression in *parentheses* and simplifying, we get:

$2_{2−(3x+2)}(2_{−3x}) 2_{2−3x−2)}(2_{−3x}) 2_{−3x}(2_{−3x}) $

Things certainly cleared up through substitution and simplifying! Anything divided by itself must be $1$.

If you felt stuck upon doing the first steps of UnCLES, you could try plugging in your own number for $x$ as a solution strategy. If you chose $2$, then $a$ would equal $8$. Plugging in $4$ for $x$ and $8$ for $a$ in the fraction, you’d arrive at $2_{−6}2_{−6} $, which is $1$. As you might guess from the fact that $1$ is the answer, it doesn’t matter what you choose to plug in for $x$; $1$ will always be the result!

All rights reserved ©2016 - 2024 Achievable, Inc.