### All SAT Math Resources

## Example Questions

### Example Question #1 : Exponents

If (300)(400) = 12 * 10* ^{n}*,

*n*=

**Possible Answers:**

2

4

3

7

12

**Correct answer:**

4

(300)(400) = 120,000 or 12 * 10^{4}.

### Example Question #2 : Exponents

(2x10^{3}) x (2x10^{6}) x (2x10^{12}) = ?

**Possible Answers:**

6x10^{23}

6x10^{21}

8x10^{23}

8x10^{21}

**Correct answer:**

8x10^{21}

The three two multiply to become 8 and the powers of ten can be added to become 10^{21}.

### Example Question #3 : Exponential Operations

If 3^{x} = 27, then 2^{2x }= ?

**Possible Answers:**

32

64

3

9

8

**Correct answer:**

64

- Solve for x in 3
^{x}= 27. x = 3 because 3 * 3 * 3 = 27. - Since x = 3, one can substitute x for 3 in 2
^{2x } - Now, the expression is 2
^{2*3} - This expression can be interpreted as 2
^{2 * }2^{2 }* 2^{2}. Since 2^{2 }= 4, the expression can be simplified to become 4 * 4 * 4 = 64. - You can also
*multiply the powers*to simplify the expression. When you*multiply the powers*, you get 2^{6}, or 2 * 2 * 2 * 2 * 2 * 2 - 2
^{6 }= 64.

### Example Question #4 : Exponential Operations

Find the value of x such that:

8^{x-3} = 16^{4-x}

**Possible Answers:**

7/2

25/7

11/3

19/4

4

**Correct answer:**

25/7

In order to solve this equation, we first need to find a common base for the exponents. We know that 2^{3} = 8 and 2^{4} = 16, so it makes sense to use 2 as a common base, and then rewrite each side of the equation as a power of 2.

8^{x-3} = (2^{3})^{x-3}

We need to remember our property of exponents which says that (a^{b})^{c} = a^{bc}.

Thus (2^{3})^{x-3 }= 2^{3(x-3)} = 2^{3x - 9}.

We can do the same thing with 16^{4-x}.

16^{4-x} = (2^{4})^{4-x} = 2^{4}^{(4-x) }= 2^{16-4x}.

So now our equation becomes

2^{3x - 9 }= 2^{16-4x}

In order to solve this equation, the exponents have to be equal.

3x - 9 = 16 - 4x

Add 4x to both sides.

7x - 9 = 16

Add 9 to both sides.

7x = 25

Divide by 7.

x = 25/7.

### Example Question #1 : Exponential Operations

Which of the following is equal to 4^{10} + 4^{10} + 4^{10} + 4^{10} + 4^{11}?

**Possible Answers:**

2^{23}

2^{15}

2^{50}

2^{40}

2^{60}

**Correct answer:**

2^{23}

We can start by rewriting 4^{1}^{1} as 4 * 4^{10}. This will allow us to collect the like terms 4^{10} into a single term.^{}

4^{10} + 4^{10} + 4^{10} + 4^{10} + 4^{11}

= 4^{10} + 4^{10} + 4^{10} + 4^{10} + 4 * 4^{10}

= 8 * 4^{10}

Because the answer choices are written with a base of 2, we need to rewrite 8 and 4 using bases of two. Remember that 8 = 2^{3}, and 4 = 2^{2}.

8 * 4^{10}

= (2^{3})(2^{2})^{10}

We also need to use the property of exponents that (a^{b})^{c} = a^{bc}. We can rewrite (2^{2})^{10} as 2^{2x10} = 2^{20}.

(2^{3})(2^{2})^{10}

= (2^{3})(2^{20})

Finally, we must use the property of exponents that a^{b }* a^{c} = a^{b+c}.

(2^{3})(2^{20}) = 2^{23}

The answer is 2^{23}.

### Example Question #6 : Exponential Operations

If 3 + 3^{n+3} = 81, what is 3^{n+2 }?

**Possible Answers:**

9

3

81

26

18

**Correct answer:**

26

3 + 3^{n+3} = 81

In this equation, there is a common factor of 3, which can be factored out.

Thus, 3(1 + 3^{n+2}) = 81

Note: when 3 is factored out of 3^{n+3}, the result is 3^{n+2} because (3^{n+3} = 3^{1} * 3^{n+2}). Remember that exponents are added when common bases are multiplied. Also remember that 3 = 3^{1}.

3(1 + 3^{n+2}) = 81

(1 + 3^{n+2}) = 27

3^{n+2} = 26

Note: do not solve for n individually. But rather seek to solve what the problem asks for, namely 3^{n+2}.

### Example Question #7 : Exponential Operations

If f(x) = (2 – x)^{(x/3)}, and 4^{n} = f(10), then what is the value of n?

**Possible Answers:**

2

5

–5

0

–2

**Correct answer:**

5

First, let us use the definiton of f(x) to find f(10).

f(x) = (2 – x)^{(x/3)}

f(10) = (2 – 10)^{(10/3)}

= (–8)^{(10/3)}

In order to evaluate the above expression, we can make use of the property of exponents that states that a^{bc} = (a^{b})^{c} = (a^{c})^{b}.

(–8)^{(10/3)} = (–8)^{10(1/3)} = ((–8)^{(1/3)})^{10}.

(–8)^{(1/3)} requires us to take the cube root of –8. The cube root of –8 is –2, because (–2)^{3} = –8.

Let's go back to simplifying ((–8)^{(1/3)})^{10}.

((–8)^{(1/3)})^{10} = (–2)^{10} = f(10)

We are asked to find n such that 4^{n} = (–2)^{10}. Let's rewrite 4^{n} with a base of –2, because (–2)^{2} = 4.

4^{n} = ((–2)^{2})^{n} = (–2)^{2n} = (–2)^{10}

In order to (–2)^{2n} = (–2)^{10}, 2n must equal 10.

2n = 10

Divide both sides by 2.

n = 5.

The answer is 5.

### Example Question #8 : Exponential Operations

What is the value of n that satisfies the following equation?

2* ^{n}*·4

*·8*

^{n}*·16 = 2*

^{n}*·4*

^{-n}*·8*

^{-}^{n}

^{-}^{n}**Possible Answers:**

2/3

–1/3

0

1/3

–2/3

**Correct answer:**

–1/3

In order to solve this equation, we are going to need to use a common base. Because 2, 4, 8, and 16 are all powers of 2, we can rewrite both sides of the equation using 2 as a base. Since 2^{2} = 4, 2^{3} = 8, and 2^{4} = 16, we can rewrite the original equation as follows:

2* ^{n} * *4

*8*

^{n}**16 = 2*

^{n}**4*

^{–}^{n }**8*

^{–}^{n }*

^{–}^{n}2* ^{n}*(2

^{2})

*(2*

^{n}^{3})

*(2*

^{n}^{4}) = 2

*(2*

^{–}^{n}^{2})

^{–}*(2*

^{n}^{3})

^{–}

^{n}Now, we will make use of the property of exponents which states that (*a ^{b}*)

*=*

^{c}*a*.

^{bc}2* ^{n}*(2

^{2n})(2

^{3n})(2

^{4}) = 2

*(2*

^{–}^{n}

^{–}^{2n})(2

^{–}^{3n})

Everything is now written as a power of 2. We can next apply the property of exponents which states that *a ^{b}a^{c}* =

*a*

^{b+c}.

2^{(n+2n+3n+4)} = 2^{(–n + –2n + –3n)}

We can now set the exponents equal and solve for *n*.

*n *+ 2*n *+ 3*n *+ 4 = –*n* + –2*n* + –3*n*

Let's combine the *n*'s on both sides.

6*n* + 4 = –6*n*

Add 6*n* to both sides.

12*n* + 4 = 0

Subtract 4 from both sides.

12*n* = –4

Divide both sides by 12.

*n* = –4/12 = –1/3

The answer is –1/3.

### Example Question #1 : Exponential Operations

If 125^{2x–4} = 625^{7–x}, then what is the largest prime factor of *x*?

**Possible Answers:**

2

5

3

7

11

**Correct answer:**

2

First, we need to solve 125^{2x–4} = 625^{7–x }. When solving equations with exponents, we usually want to get a common base. Notice that 125 and 625 both end in five. This means they are divisible by 5, and they could be both be powers of 5. Let's check by writing the first few powers of 5.

5^{1} = 5

5^{2} = 25

5^{3} = 125

5^{4} = 625

We can now see that 125 and 625 are both powers of 5, so let's replace 125 with 5^{3} and 625 with 5^{4}.

(5^{3})^{2x–4} = (5^{4})^{7–x}

Next, we need to apply the rule of exponents which states that (*a ^{b}*)

*=*

^{c}*a*.

^{bc}5^{3(2x–4)} = 5^{4(7–x)}

We now have a common base expressed with one exponent on each side. We must set the exponents equal to one another to solve for *x*.

3(2*x –* 4) = 4(7 – *x*)

Distribute the 3 on the left and the 4 on the right.

6*x* – 12 = 28 – 4*x*

Add 4*x* to both sides.

10*x* – 12 = 28

Add 12 to both sides.

10*x* = 40

Divide by 10 on both sides.

*x* = 4

However, the question asks us for the largest prime factor of *x*. The only factors of 4 are 1, 2, and 4. And of these, the only prime factor is 2.

The answer is 2.

### Example Question #10 : Exponential Operations

(*x*^{3})^{2} * *x ^{–}*

^{2}=

**Possible Answers:**

*x*^{4}

*x*^{2}

*x*

*x*^{6}

*x ^{–}*

^{4}

**Correct answer:**

*x*^{4}

When an exponent is raised to a power, we multiply. But when two exponents with the same base are multiplied, we add them. So (*x*^{3})^{2} = *x*^{3*2} = *x*^{6}. Then (*x*^{3})^{2} * *x ^{–}*

^{2}=

*x*

^{6}*

*x*

^{–}^{2}=

*x*

^{6 – 2}=

*x*

^{4}.

Certified Tutor