PSAT Math - How to add exponents

If _a_2 = 35 and _b_2 = 52 then _a_4 + _b_6 = ?

141,833

522

3929

140,608

150,000

Explanation

_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

If $3^{4x+6}=27^{2x}$ , what is the value of ?

Explanation

Using exponents, 27 is equal to 33. So, the equation can be rewritten:

34_x_ + 6 = (33)2_x_

34_x_ + 6 = 36_x_

When both side of an equation have the same base, the exponents must be equal. Thus:

4_x_ + 6 = 6_x_

6 = 2_x_

x = 3

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

Explanation

Since we have two ’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 : $9^{ (x + 5)} = 81 = 9^ 2$

Thus or

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

How many of the following base ten numbers have a base five representation of exactly four digits?

(A)

(B)

(C)

(D)

Three

Four

None

One

Two

Explanation

A number in base five has powers of five as its place values; starting at the right, they are $1, 5 ^{1} = 5, 5 ^{2} = 25, 5 ^{3} = 125,5 ^{4} = 125$

The lowest base five number with four digits would be

$1000 _{\textrm{five}} = 1 \times 5^{3} = 125$ in base ten.

The lowest base five number with five digits would be

$10000 _{\textrm{five}} = 1 \times 5^{4} = 625$ in base ten.

Therefore, a number that is expressed as a four-digit number in base five must fall in the range

$125 \leq N \leq 624$

Three of the four numbers - all except 100 - fall in this range.

Evaluate:

$-\left ( -3 \right )^{0}-\left ( -3^{0} \right )$

Explanation

$-\left ( -3 \right )^{0}= -1$

$(-3^{0})=-1$

Solve for x.

23 + 2x+1 = 72

Explanation

The answer is 5.

8 + 2x+1 = 72

2x+1 = 64

2x+1 = 26

x + 1 = 6

x = 5

If $\dpi{100} \small r$ and $\dpi{100} \small s$ are positive integers, and $\dpi{100} \small 25\left ( 5^{r} \right )=5^{s-2}$ , then what is $\dpi{100} \small s$ in terms of $\dpi{100} \small r$ ?

$\dpi{100} \small r+4$

$\dpi{100} \small r$

$\dpi{100} \small r+1$

$\dpi{100} \small r+2$

$\dpi{100} \small r+3$

Explanation

$\dpi{100} \small 25\left ( 5^{r} \right )$ is equal to $5^{2}\left ( 5^{r}\right )$ which is equal to $\dpi{100} \small \left ( 5^{r+2} \right )$ . If we compare this to the original equation we get $\dpi{100} \small r+2=s-2\rightarrow s=r+4$

What is the value of such that ?

Explanation

We can solve by converting all terms to a base of two. 4, 16, and 32 can all be expressed in terms of 2 to a standard exponent value.

$4=2^2\rightarrow 4^2=(2^2)^2$

$16=2^4\rightarrow 16^3=(2^4)^3$

$32=2^5\rightarrow 32^2=(2^5)^2$

We can rewrite the original equation in these terms.

Simplify exponents.

$2^n=(2^4)*(2^{12})*(2^{10})$

Finally, combine terms.

$2^n=2^{4+12+10}=2^{26}$

From this equation, we can see that .

Simplify:

$3\sqrt{18} + 2\sqrt{50}$

$19\sqrt{2}$

$9\sqrt{2}$

$10\sqrt{2}$

$5\sqrt{68}$

$None\ of\ the\ above$

Explanation

$3\sqrt{18} = 3\sqrt{2\star 3^{2}} = 9\sqrt{2}$

Similarly $2\sqrt{50} = 2\sqrt{2\star 5^{2}} = 10\sqrt{2}$

So $9\sqrt{2} + 10\sqrt{2}= 19\sqrt{2}$

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?

1/5

5_b–_1

Explanation

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_b +(–5(_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_b_ + (–1)(5(b–1))(5) = 5_b_ – (5(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, abac = a b+c. We can rewrite 5 as 51 and then apply this rule.

5_b_ – (5(_b–1))(5) = 5_b – (5(_b–1))(51) = 5_b – 5(_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_b – 5_b = 0

The answer is 0.

How to add exponents

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