Textbook

Exponents can make a simple problem seem hard. However, with a clear understanding of the exponent rules, these problems can be much simpler than they look.

Here are the exponent rules that you will need to know.

$3_{4}∗3_{5}=3_{9}$

In this problem, both bases of $3_{4}$ and $3_{5}$ are $3$. When you expand both exponents you get the following:

$3_{4}×3_{5}=(3×3×3×3)×(3×3×3×3×3)=3_{9}$

Notice how both exponents expand into many threes written in a row. This problem is just $3$ multiplied by itself $9$ separate times. The exponents, being $4$ and $5$, have a sum of $4+5=9$.

Note that you can only add exponents if the base numbers are the same!

$3_{6}/3_{3}=3_{3}$

Like before, the bases of $3_{6}$ and $3_{3}$ are $3$. Thus, we can subtract the exponents. The subtraction is essentially the cancellation of three of the $3$s on the top. The exponents, being $6$ and $3$, have a difference of $6−3=3$.

$3_{6}/3_{3}=(3×3×3×3×3×3)/(3×3×3)=3×3×3=3_{3}$

Try this GRE problem that involves Rule 1 and 2:

$0<x<1$

Quantity A: $2_{4}(2_{3})x_{3}/2_{5}$

Quantity B: $4_{4}x_{5}/(4x)_{3}$

Do you know the answer?

(spoiler)

Answer: *Quantity B is greater*

Let’s simplify Quantity A first:

$2_{4}(2_{3})x_{3}/2_{5}2_{7}x_{3}/2_{5}2_{2}x_{3}4x_{3} $

And let’s simplify Quantity B:

$4_{4}x_{5}/(4x)_{3}4_{4}x_{5}/((4_{3})(x_{3}))4_{4}x_{2}/4_{3}4_{1}x_{2}4x_{2} $

Now we can compare them side by side:

$A:B: 4x_{3}4x_{2} $

Given that the question states $0<x<1$, Quantity B must be greater. A fraction multiplied by itself shrinks, and a fraction multiplied by itself more will shrink more.

$3_{−4}=1/(3_{4})$

Simply flip the fraction and rewrite the exponent without the negative in front. Notice how a negative exponent does not describe a negative number. Remembering **Rule 2**, we can see that $3_{4}/3_{8}$ would be equal to $3_{−4}$ because $4−8=−4$. Because four of the $3$s on the bottom are canceled out by the four of the $3$s on the top, we are left with $1/(3_{4})$.

$3_{0}=1$

Remembering **Rule 2**, any exponented number minus that same number is $0$. When we subtract equal exponents, we have an exponent of $0$. Looking at it as a fraction, we can also see that everything on the top is equivalent to everything on the bottom. When the numerator equals the denominator, the fraction is always equal to $1$.

$3_{3}/3_{3}=3_{3−3}=3_{0}=1$

Try this GRE question that involves Rule 3 and 4:

Which of the following represents $y$ in terms of $z$?

Given: $z_{−4}z_{4}x_{−y}=x_{y}x_{z}$

A. $z$

B. $z/2$

C. $−z/2$

D. $2/z$

E. $−2/z$

Try to solve it, and check your work!

(spoiler)

Answer: *C. $−z/2$*

Let’s simplify the expression step by step.

$z_{−4}z_{4}x_{−y}z_{−4+4}x_{−y}z_{0}x_{−y}(1)x_{−y}x_{−y}1/x_{y}11/x_{z}x_{−z}−z−z/2 =x_{y}x_{z}=x_{y}x_{z}=x_{y}x_{z}=x_{y}x_{z}=x_{y}x_{z}=x_{y}x_{z}=x_{2y}x_{z}=x_{2y}=x_{2y}=2y=y $

The way above illustrates how to use negative exponents to manipulate equations. However, you could have also simplified it in other ways, like this:

$z_{−4}z_{4}x_{−y}z_{−4+4}x_{−y}z_{0}x_{−y}x_{−y}x_{−y}−y0−z−z/2 =x_{y}x_{z}=x_{y}x_{z}=x_{y}x_{z}=x_{y}x_{z}=x_{y+z}=y+z=2y+z=2y=y $

No matter which approach you take, you’ll get the same answer!

$(3_{3})_{2}=3_{6}$

This can be rewritten as $(3×3×3)×(3×3×3)$. Because there are six $3$s multiplied in a row, this is equivalent to $3_{6}$. Incidentally, if you multiply the exponents, you will get the same answer for the exponent $3×2=6$.

$3_{3}×4_{3}=12_{3}$

This problem is essentially $(3×3×3)×(4×4×4)$. If you combine the pairs of $3$s and $4$s, you will get three $12$s, i.e.: $12×12×12=12_{3}$.

Try this GRE question that involves Rule 5 and 6:

Which mixed number represents this fraction?

$((2_{2})_{2})_{2}3_{2}11_{2} $

A. $325625 $

B. $425665 $

C. $4256155 $

D. $4256235 $

E. $525665 $

Give it a try and check your work!

(spoiler)

Answer: *B. $425665 $*

Let’s simplify! We’ll write the fraction horizontally so it’s easier to follow:

$3_{2}11_{2}/((2_{2})_{2})_{2}(3∗11)_{2}/((2_{2})_{2})_{2}33_{2}/((2_{2})_{2})_{2}1089/((2_{2})_{2})_{2}1089/(2_{2})_{2∗2}1089/(2_{2})_{4}1089/2_{2∗4}1089/2_{8}1089/256≈4.25390625 $

We’ve simplified the expression, now what? There’s a few ways we can take it from here. To start, our final value is $4+0.25390625$, so only the choices with a $4$ whole number could be correct, eliminating choices *A* and *E*. Looking at just the fractional part $4+0.253906254$, we can see that it is less than $0.5$, which eliminates choices *C* and *D* since their fractional parts are greater than $0.5$. That leaves us with choice *B* as the correct answer.

If you want to double-check (and you should to help avoid mistakes), you can plug in the fractional parts into your calculator. You’ll see that $65/256=0.25390625$, confirming we’ve made the correct choice.

It can be useful to translate radicals into exponents and use your exponent rules to solve for the answer. Remember, a radical can be described as a fraction exponent by flipping the radical root and inner exponent:

$25 =225_{1} =25_{1/2}$

Here’s a video going through one of our practice questions to demonstrate these ideas in action:

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