# The Power of Exponents on the GMAT

The GMAT Quantitative section gives you 75 minutes to answer 37 questions. While a little over two minutes per question may sound generous, it’s going to feel anything but on test day.

So particularly if you haven’t had a math class for a while, it’s important that you master some basic concepts that commonly appear on the exam—one of which is working with exponents. Below, I’ll talk about the most common rules that you should know about exponents and finish with some examples that you can try on your own.

## What Are Exponents?

Let’s start out with the basics. When you see an expression of the form by, b is known as the base and y is the exponent. The meaning of the expression is to multiply the base by itself the number of times given by the exponent.

For example:

b2 = bb           or          22 = 2 ∙ 2 = 4

b3 = bbb      or          23 = 2 ∙ 2 ∙ 2 = 8

Exponents can be any number–i.e. positive, negative, 0, integer, fraction, etc. You probably won’t encounter all of these on your GMAT, but I’ll cover the main rules that you should know.

## The Rules of Exponents

 Rule Description Expression Example 1. Power of zero rule Any number raised to the 0 power is 1. x0 = 1 20 = 1 2. Power of one rule Any number raised to the 1st power is that number. x1 = x 31 = 3 3. Product rule Multiplying a common base with different exponents is equivalent to adding the exponents (ym) ∙ (yn) = y(m+n) (22) ∙ (23) = 2(5) = 32 4. Power rule Raising an exponential expression to another exponent is equivalent to multiplying the exponents. (ym)n = y(m ∙ n) (22)3 = 26 = 64 5. Quotient rule Dividing a common base with different exponents is equivalent to subtracting the exponents. (ym) / (yn) = y(m – n) (35) / (32) = 33 = 27 6. Expanded power rule Raising the product of two numbers to a power is equivalent to the product of each of those numbers individually raised to that power (x ∙ y)n = xn ∙ yn (3 ∙ 4)3 = 33 ∙ 43 = 1728 7. Reciprocal rule A reciprocal is what to multiply by a number to get one. When a base is raised to a negative power, reciprocate (find the reciprocal of) the base, keep the exponent with the original base, and drop the negative. y-1 = 1/y 2-1 = ½ 8. Negative exponent rule See reciprocal rule above. y–m = 1/(ym) 3-3 = 1/(33) = 1/27 9. Expanded negative exponent rule See reciprocal rule above. (y/z)–n = zn / yn (2/3)-4 = 34 / 24= 81/16 10. Complex expression exponent equality rule If a common base b raised to the x power equals the base b raised to the y power, then x = y. If bx = by, then x = y 3(2∙y) = 34 2 ∙ y = 4 y  = 2 11. Fractional exponent rules The nth root of y = y(1/n) and the nth root of a base to an exponent m is simplified as the nth root of ym = y(m/n). The point to keep in mind is that square and square root are inverses as are cube and cube root. n√y = y1/n 3√27 = 271/3  = 3 12. Scientific notation You should be able to routinely shift back and forth between fraction, decimal, and scientific notation. a × 10k   where k is an integer and 1 ≤ |a| ≤ 10 For numbers n ≥ 1 100 = 1 ∙ (102), 89 = 8.9 ∙ (101) 7,354 = 7.354 ∙ (103). For numbers n < 1 0.01 = 1 ∙ (10-2), 0.89 = 8.9 ∙ (10-1) 0.007354 = 7.354 ∙ (10-3). 13. Factoring Factoring is a subject on its own but is often closely linked to exponents, so you should know the two key examples 1. (y + z)2 = y2 ++ 2yz + z2 2. (y –z)(y +z) = y2 – z2 1. (3 + 2)2 = 32 + 2 ∙ 3 ∙ 2 + 22 = 25 2. (3 – 2)(3 + 2) = 32 – 22 = 5

Let’s practice how to use each of these rules, alone or in combination, in the examples below. These examples are similar to GMAT problems that you will encounter in the Problem Solving and Data Sufficiency sections.

## Examples to Work Through

1) If a and b are positive integers and (2a)b = 23, what is the value of 2a ∙ 2b?

(A)         6

(B)         8

(C)         16

(D)         32

(E)         64

Explanation: (2a)b = 2(a b) = 23 by rule 4 above. Thus, ab = 3. The integer solution is where either a or b is 1 and the remaining variable is 3. Thus (2a) ∙ (2b) = 2(a+b) => a + b = 3 + 1 = 4 and 24 = 16. Thus, the answer is (C).

2) 3/100 + 5/1,000 + 7/100,000 =

(A)         0.357

(B)         0.3507

(C)         0.35007

(D)         0.0357

(E)         0.03507

Explanation: Applying rule 12 above, 3/100 = 0.03, 5/1000 = 0.005, and 7/100,000 = 0.00007. Thus, the answer is (E).

3) In which of the following pairs are the two numbers reciprocals of each other?

I. 3 and 1/3

II. 1/17 and -1/17

III.  √3 and (√3)/3

(A) I only

(B) II only

(C) I and II

(D) I and III

(E) II and III

Explanation: Two numbers that multiply together to give you one are reciprocals.

3 × 1/3 = 1 so choice I gives a pair of reciprocals.

1/17 × -1/17 = – 1/289  so this is NOT a pair of reciprocals.

√3 x (√3)/3 = 1 (√3 x √3)/3 = 3/3 = 1 so choice III also gives a pair of reciprocals.

4) 173 + 174 =

(A)         177

(B)         173(18)

(C)         176(18)

(D)         2(173) + 17

(E)         2(173) – 17

Explanation: Factor out 173 and simplify.

173 + 174 =

173 + 173+1 =

173 + 173171 =

173 (1 + 17) =

173 (18)

5) (1 +√5 )(1 – √5) =

(A)         -4

(B)         2

(C)         6

(D)         -4 – 2√5

(E)         6 – 2√5

Explanation: Apply the factoring rule 13 above. Doing the multiplication yields 12 – √5 2 = 1 – 5 = -4. Thus, the answer is (A).

6) Of the following, the closest approximation to √(5.98(601.5))/15.79  is

(A)         5

(B)         15

(C)         20

(D)         25

(E)         225

Explanation: Use nearby numbers to approximate.

(6 ∙ 600)/16 =

3,600/16 =

225

√225 = 15.

7) What is the largest integer, n, such that 1/(2n) > 0.01?

(A)         5

(B)         6

(C)         7

(D)         10

(E)         51

Explanation: Try some values as examples. If we pick n = 5, then 1 / (25) = 1/32 = 0.03.

Try the next value 6. This yields around 0.015, so going any further will fail.

You can try  ≈ 0.0078 to be sure. Thus, the answer is (B).

8) If x and k are integers and (12x)(42x+1) = (2k)(32), what is the value of k?

(A)         5

(B)         7

(C)         10

(D)         12

(E)         14

Explanation: Think about how to write (12x)(42x+1) in terms of (2k)(32). To do this, use rule 10 above.

12x = (3∙ 22)x = 3x ∙ 22x and 42x+1 = (22)2x+1 = 22(2x+1) = 24x+2

By substitution we get: 12x42x+1 = 3x ∙ 22x ∙ 24x+2

Now we can combine the base 2.

12x42x+1 = 3x ∙ 22x+4x+2 = 3x ∙ 26x+2

Now we can write:

3x ∙ 26x+2 = 32 ∙ 2k and 3x = 32 so x = 2. Then, k = 6x + 2 = 6(2) + 2 = 14.

9) If r and s are positive integers such that (2r)(4s) = 16, then 2r + s =

(A)         2

(B)         3

(C)         4

(D)         5

(E)         6

Explanation: Apply rules 10 and 13.

2r ∙ 4s = 2r ∙ 22s =2(r+2s) and set this equal to 24

So, because the exponents are equal, r + 2s and 4 equal 4. This implies r = 2 and s = 1 because r and s are positive integers. Thus, 2r + s =2(2)+1= 5. The answer is (D).

For the data sufficiency questions 10-17 below, the possible answers are the following:

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are NOT sufficient.

10) If n is positive, is √n > 100?

(1) √(n – 1) > 99

(2) √(n + 1) > 101

Explanation:

Square both sides on both the original problem and the choices.

√n > 10,000

This means you need a condition with a value of n greater than 10,000.

n – 1 > 9,801 implies n > 9,802, so (1) is not sufficient.

n + 1 > 10201 implies n > 10,200

and

10,200 > 10,000

so

√n > 100.

Thus, (2) alone is sufficient and the answer is (B).

11) If x ≠ 0, what is the value of ( xp / xq )4?

(1) p = q

(2) x = 3

Explanation: If p = q, then the value of ( xp / xq ) is 1, so condition (1) is sufficient.

If x = 3, then the expression could take many values depending on the values of p and q.

12) If x2 + y2 = 29 has only one possible solution, what is the value of (xy)2?

(1) xy = 10

(2)  x = 5

Explanation: (x + y)2 = x2 + 2xy + y2 by rule 13.

(xy)2 = x2 – 2xy + y2.

You can then say x2 + y2 = 29 and -2xy = -20.

So if you know xy, then you have sufficient information to evaluate (xy)2.

13) If x > 0, what is the value of x5?

(1) √x = 32

(2) x2 = 220

Explanation: Squaring both sides of condition (1), gives you x = 3232 so x5 is knowable and condition (1) is sufficient.

For condition (2), if you know x2, then you can determine x5 by raising x2 to the exponent 5/2.

Thus, either alone is sufficient and the answer is (D).

14) What is the value of the positive integer n?

(1) n4 < 25

(2) nn2

Explanation: Condition (1) limits your choices to 1 or 2 because 14 < 25 and 24 < 25 and these are the only integers that when raised to the 4th power will have a value less than 25.

However this is not sufficient.

(2) eliminates the possible value of n= 1, so the answer is (C). You need both conditions to determine the value of n.

15) If x is a positive integer, is  an integer?

(1) √4x is an integer

(2) √3x is not an integer

Explanation: √4x = 2√x by rule 6. The integers are closed under multiplication, so if 2 times √x  is an integer, then √x must also be an integer, so condition (1) is sufficient.

For condition (2), we are told the solution is not an integer. It is possible to choose an integer value for x and a non-integer value for x so that √3x is not an integer (for example consider x = 5 and x = π.

So this condition does not help answer the question. Thus, the best answer is (A).

16) If x is a positive integer, what is the value of x?

(1) x2√x

(2)  n/x = n and n ≠ 0

Explanation: Square both sides of condition (1), so x4 = x.

This means x = 0 or x = 1.

We want positive values only, so this is sufficient to state that x = 1.

For condition (2), it follows that n = nx  so x = 1.

This means condition (2) is also sufficient. Thus, the answer is (D).

17) If a and b are positive integers, is 3√ab an integer?

(1) √a is an integer

(2) b√a

Explanation: For condition (1) you can pick some examples.

Pick a = 4 and b = 5.

√4 = 2 so √a  is an integer, but 20(1/3) is not an integer (3√ab).

So condition (1) is not sufficient.

As for condition (2), we can show:

3√ab given b√a .

3√ab = 3√(a√a) = (a ∙ a(1/2))(1 /3)

(a1 ∙ a(1/2))(1 /3) = (a(2/2) ∙ a(1/2))  = (a(3/2))(1 /3)

If b = √a, then (ab) = (a(3/2))

(a(3/2))(1 /3) a(1/2) = √a = b

But we know that b is an integer, so this is sufficient. Thus, the answer is (B).
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