### All GRE Math Resources

## Example Questions

### Example Question #1 : Exponential Operations

Simplify

**Possible Answers:**

None

**Correct answer:**

Divide the coefficients and subtract the exponents.

### Example Question #12 : How To Divide Exponents

Which of the following is equal to the expression , where

xyz ≠ 0?

**Possible Answers:**

z

xyz

xy

z/(xy)

1/y

**Correct answer:**

1/y

(xy)^{4} can be rewritten as x^{4}y^{4} and z^{0} = 1 because a number to the zero power equals 1. After simplifying, you get 1/y.

### Example Question #2 : How To Divide Exponents

If , then

**Possible Answers:**

Cannot be determined

**Correct answer:**

Start by simplifying the numerator and denominator separately. In the numerator, (c^{3})^{2} is equal to c^{6}. In the denominator, c^{2 }* c^{4} equals c^{6} as well. Dividing the numerator by the denominator, c^{6}/c^{6}, gives an answer of 1, because the numerator and the denominator are the equivalent.

### Example Question #1 : How To Divide Exponents

If , which of the following is equal to ?

**Possible Answers:**

The answer cannot be determined from the above information

a^{4}

a^{6}

a

a^{18}

**Correct answer:**

a^{18}

The numerator is simplified to (by adding the exponents), then cube the result. a^{24}/a^{6} can then be simplified to .

### Example Question #2 : Exponential Operations

[64^{1/2} + (–8)^{1/3}] * [4^{3}/16 – 3^{171}/3^{169}] =

**Possible Answers:**

30

–30

–5

9

16

**Correct answer:**

–30

Let's look at the two parts of the multiplication separately. Remember that (–8)^{1/3} will be negative. Then 64^{1/2} + (–8)^{1/3 }= 8 – 2 = 6.

For the second part, we can cancel some exponents to make this much easier. 4^{3}/16 = 4^{3}/4^{2} = 4. Similarly, 3^{171}/3^{169} = 3^{171–169} = 3^{2} = 9. So 4^{3}/16 – 3^{171}/3^{169 }= 4 – 9 = –5.

Together, [64^{1/2} + (–8)^{1/3}] * [4^{3}/16 – 3^{171}/3^{169}] = 6 * (–5) = –30.

### Example Question #1 : Exponential Operations

Evaluate:

**Possible Answers:**

**Correct answer:**

Distribute the outside exponents first:

Divide the coefficient by subtracting the denominator exponents from the corresponding numerator exponents:

### Example Question #2 : Exponential Operations

**Possible Answers:**

**Correct answer:**

The easiest way to solve this is to simplify the fraction as much as possible. We can do this by factoring out the greatest common factor of the numerator and the denominator. In this case, the GCF is .

Now, we can cancel out the from the numerator and denominator and continue simplifying the expression.

### Example Question #3 : Exponential Operations

Simplify: y^{3}x^{4}(yx^{3} + y^{2}x^{2} + y^{15} + x^{22})

**Possible Answers:**

y^{3}x^{12} + y^{12}x^{8} + y^{24}x^{4} + y^{3}x^{23}

y^{4}x^{7} + y^{5}x^{6} + y^{18}x^{4} + y^{3}x^{26}

y^{3}x^{12} + y^{6}x^{8} + y^{45}x^{4} + y^{3}x^{88}

y^{3}x^{12} + y^{6}x^{8} + y^{45} + x^{88}

2x^{4}y^{4} + 7y^{15} + 7x^{22}

**Correct answer:**

y^{4}x^{7} + y^{5}x^{6} + y^{18}x^{4} + y^{3}x^{26}

When you multiply exponents, you add the common bases:

y^{4} x^{7} + y^{5}x^{6} + y^{18}x^{4} + y^{3}x^{26}

### Example Question #1 : How To Add Exponents

Indicate whether Quantity A or Quantity B is greater, or if they are equal, or if there is not enough information given to determine the relationship.

Quantity A:

Quantity B:

**Possible Answers:**

Quantity A is greater.

Quantity B is greater.

The quantities are equal.

The relationship cannot be determined from the information given.

**Correct answer:**

Quantity B is greater.

By using exponent rules, we can simplify Quantity B.

^{}

Also, we can simplify Quantity A.

^{}

Since n is positive,

### Example Question #4 : Exponential Operations

If , what is the value of ?

**Possible Answers:**

**Correct answer:**

Rewrite the term on the left as a product. Remember that negative exponents shift their position in a fraction (denominator to numerator).

The term on the right can be rewritten, as 27 is equal to 3 to the third power.

Exponent rules dictate that multiplying terms allows us to add their exponents, while one term raised to another allows us to multiply exponents.

We now know that the exponents must be equal, and can solve for .