Evaluating Powers, Including Negative Bases | Grade 8 Math | LevelKite

A power is a short way to write a product that uses the same number as a factor again and again. In the power $5^3$ , the base $5$ is the number being multiplied, and the exponent $3$ tells you how many times the base is used as a factor. So $5^3 = 5 \cdot 5 \cdot 5$ . To evaluate a power means to find its value.

In general, let the base $b$ be any number (a negative base is written in parentheses), and let the exponent $n$ be a positive whole number, such as $1$ , $2$ or $3$ . Every exponent in this lesson is a positive whole number. Then $b^n$ means $n$ factors of $b$ multiplied together:

b^n = \underbrace{b \cdot b \cdot \cdots \cdot b}_{n \text{ factors}}

Worked example

A positive base

Evaluate $3^4$ .

The exponent is 4, so write the base 3 as a factor four times. That is exactly what the power means.
$3^4 = 3 \cdot 3 \cdot 3 \cdot 3$
Multiply the first two factors: 3 times 3 is 9. Replacing part of a product with its value does not change the whole product.
$3 \cdot 3 \cdot 3 \cdot 3 = 9 \cdot 3 \cdot 3$
Multiply the product so far by the next factor: 9 times 3 is 27. Again, the value of the whole product stays the same.
$9 \cdot 3 \cdot 3 = 27 \cdot 3$
Multiply by the last factor: 27 times 3 is 81. All four factors of 3 are now used, so 81 is the value of the power.
$27 \cdot 3 = 81$

Answer

The value of $3^4$ is $81$ .

The base can also be a negative number. To show that $-2$ is the base, write it inside parentheses: $(-2)^3 = (-2)(-2)(-2)$ , which is three factors of $-2$ . The table lists the first six powers of $-2$ .

Powers of negative 2

Power	Written as factors	Exponent is	Value
$(-2)^1$	$-2$	odd	$-2$
$(-2)^2$	$(-2)(-2)$	even	$4$
$(-2)^3$	$(-2)(-2)(-2)$	odd	$-8$
$(-2)^4$	$(-2)(-2)(-2)(-2)$	even	$16$
$(-2)^5$	$(-2)(-2)(-2)(-2)(-2)$	odd	$-32$
$(-2)^6$	$(-2)(-2)(-2)(-2)(-2)(-2)$	even	$64$

The signs switch back and forth, and pairing the factors shows why. You may group factors in any way when you multiply, so pair them up. Each pair, such as $(-2)(-2) = 4$ , is positive, because the product of two negative numbers is positive. When the exponent is even, every factor has a partner, so the whole product is positive. When the exponent is odd, one factor of $-2$ is left without a partner, and that one unpaired negative factor makes the whole product negative. The negative base changes only the sign of the result: the absolute values $2, 4, 8, 16, 32, 64$ are exactly the powers $2^1$ through $2^6$ .

Key concept

The sign of a power with a negative base

In a power with a positive whole number exponent $n$ , the exponent counts how many times the base is used as a factor. Now let $a$ be a positive number, so $-a$ is negative. Then $(-a)^n$ has the same absolute value as $a^n$ , and its sign depends on the count: it is positive when $n$ is even and negative when $n$ is odd. The parentheses make $-a$ the base. Without them, $-a^n$ means $-(a^n)$ , the opposite of $a^n$ .

(-a)^n = \begin{cases} a^n & \text{if } n \text{ is even} \\ -(a^n) & \text{if } n \text{ is odd} \end{cases}

Worked example

A negative base with an even exponent

Evaluate $(-3)^4$ .

The parentheses make negative 3 the base, and the exponent 4 says to use it as a factor four times.
$(-3)^4 = (-3)(-3)(-3)(-3)$
Group the factors into two pairs. You may group factors in any way when you multiply, so the value does not change.
$(-3)(-3)(-3)(-3) = \big[(-3)(-3)\big] \cdot \big[(-3)(-3)\big]$
Multiply within each pair. The product of two negative numbers is positive, so each pair equals 9.
$\big[(-3)(-3)\big] \cdot \big[(-3)(-3)\big] = 9 \cdot 9$
Multiply the two pair products. No negative factor is left over, so the result is positive, as the even exponent predicts.
$9 \cdot 9 = 81$

Answer

The value of $(-3)^4$ is $81$ .

Worked example

Substituting a negative number

Evaluate $x^3 - x^2$ when $x = -4$ .

Replace each x with negative 4, written inside parentheses. The variable x stands for negative 4, including its sign, so each exponent must apply to negative 4, not just to 4.
$x^3 - x^2 = (-4)^3 - (-4)^2$
Evaluate the first power. The first pair gives 16, and the one leftover factor of negative 4 makes the result negative.
$(-4)^3 = (-4)(-4)(-4) = 16 \cdot (-4) = -64$
Evaluate the second power. Its two negative factors form one pair, so the result is positive.
$(-4)^2 = (-4)(-4) = 16$
Replace each power with its value, then subtract. You found the powers first because exponents come before subtraction.
$(-4)^3 - (-4)^2 = -64 - 16 = -80$

Answer

When $x = -4$ , the value of $x^3 - x^2$ is $-80$ .

For a positive whole number $n$ , the power $b^n$ means $n$ factors of the base $b$ , and you evaluate it by multiplying those factors one step at a time. Parentheses decide whether a negative sign belongs to the base, so $(-3)^4 = 81$ but $-3^4 = -81$ . With a negative base, an even exponent gives a positive value and an odd exponent gives a negative value. Next, you will multiply and divide powers with the same base and find a power of a power.