### All PSAT Math Resources

## Example Questions

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

If *a*^{2} = 35 and *b*^{2} = 52 then *a*^{4} + *b*^{6} = ?

**Possible Answers:**

141,833

140,608

522

3929

150,000

**Correct answer:**

141,833

*a*^{4 }= *a*^{2} * *a*^{2} and *b*^{6}= *b*^{2} * *b*^{2 }* *b*^{2}

Therefore *a*^{4} + *b*^{6} = 35 * 35 + 52 * 52 * 52 = 1,225 + 140,608 = 141,833

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

If 9^{(x + 5)} + 3^{2(x + 5)} = 162, what is the value of x?

**Possible Answers:**

3

0

-3

1

-1

**Correct answer:**

-3

Since we have two x’s in 9^{(x + 5)} + 3^{2(x + 5)} we will need to combine the two terms.

For 3^{2(x + 5)} this can be rewritten as

(3^{2})^{ (x + 5)} = 9^{ (x + 5)}

So we have 9^{ (x + 5)} + 9^{ (x + 5)} = 162.

Or 2 (9^{ (x + 5)}) = 162

Divide this by 2: 9^{ (x + 5)} = 81 = 9^{ 2}

Thus x +5 = 2 or x = -3

*Hint: If you are really unsure, you could have plugged in the numbers and found that the first choice worked in the equation.

### Example Question #2261 : Sat Mathematics

Solve for x.

2^{3 }+ 2^{x+1 }= 72^{}

**Possible Answers:**

4

6

5

3

7

**Correct answer:**

5

The answer is 5.

8 + 2^{x+1 }= 72

2^{x+1 }= 64

2^{x+1 }= 2^{6}

^{ } x + 1 = 6

x = 5

### Example Question #3 : How To Add Exponents

Which of the following is eqivalent to 5* ^{b}* – 5

^{(b–1)}– 5

^{(b–1) }– 5

^{(b–1) }– 5

^{(b–1) }– 5

^{(b–1) }, where

*b*is a constant?

**Possible Answers:**

5^{b–1}

5^{}

1/5

0

1

**Correct answer:**

0

We want to simplify 5* ^{b}* – 5

^{(b–1) }– 5

^{(b–1) }– 5

^{(b–1) }– 5

^{(b–1) }– 5

^{(b–1) }.

Notice that we can collect the –5^{(}^{b–1)} terms, because they are like terms. There are 5 of them, so that means we can write –5^{(b–1) }– 5^{(b–1) }– 5^{(b–1) }– 5^{(b–1) }– 5^{(b–1) }as (–5^{(b–1)})5.

To summarize thus far:

5* ^{b}* – 5

^{(b–1) }– 5

^{(b–1) }– 5

^{(b–1) }– 5

^{(b–1) }– 5

^{(b–1) }= 5

*+(–5*

^{b}^{(b–1)})5

It's important to interpret –5^{(b–1) }as (–1)5^{(b–1)} because the –1 is not raised to the (*b* – 1) power along with the five. This means we can rewrite the expression as follows:

5* ^{b}* +(–5

^{(b–1)})5 = 5

*+ (–1)(5*

^{b}^{(b–1)})(5) = 5

*– (5*

^{b}^{(b–1)})(5)

Notice that 5^{(}^{b–1)} and 5 both have a base of 5. This means we can apply the property of exponents which states that, in general, *a ^{b}a^{c}* =

*a*

^{b+c}. We can rewrite 5 as 5

^{1}and then apply this rule.

5* ^{b}* – (5

^{(b–1)})(5) = 5

*– (5*

^{b}^{(b–1)})(5

^{1}) = 5

*– 5*

^{b}^{(b–1+1)}

Now, we will simplify the exponent *b* – 1 + 1 and write it as simply *b*.

5* ^{b}* – 5

^{(b–1+1) }= 5

*– 5*

^{b}*= 0*

^{b}The answer is 0.

### Example Question #2263 : Sat Mathematics

If, then what does equal?

**Possible Answers:**

**Correct answer:**

### Example Question #6 : Exponential Operations

Simplify. All exponents must be positive.

**Possible Answers:**

**Correct answer:**

Step 1:

Step 2:

Step 3: (Correct Answer):

### Example Question #7 : Exponents

Simplify. All exponents must be positive.

**Possible Answers:**

**Correct answer:**

Step 1:

Step 2:

Step 3:

### Example Question #7 : Exponential Operations

Answer must be with positive exponents only.

**Possible Answers:**

**Correct answer:**

Step 1:

Step 2: The above is equal to

### Example Question #8 : Exponential Operations

Evaluate:

**Possible Answers:**

**Correct answer:**

### Example Question #2264 : Sat Mathematics

Simplify:

**Possible Answers:**

**Correct answer:**

Similarly

So