Welcome back to our series on exponents. Today we will see what happens when we throw negatives into our exponential expressions. We will explore both negative bases and negative exponents.

First, the bases. The rule to remember for negative bases is that *odd powers of negative bases are negative and even powers of negative bases are positive*. This rule makes sense when you remember that exponents simply notate a number of multiplications by the base (and remember your rules about multiplication with negative factors).

Multiplication with an odd number of negative factors yields a negative product:

(-x)^{3} = -x * -x * -x

(-x)^{3} < 0

Multiplication with an even number of negative factors yields a positive product:

(-x)^{4} = -x * -x * -x * -x

(-x)^{4} > 0

A note on notation: parentheses should always be used around a negative value as the base of an exponent. If they are not, then the order of operations dictates that the exponent be applied before the negative sign. To avoid confusion, whenever the negative is meant to be left out of the exponential operation, parentheses are used like this: -(x)4 to make the order clear. Please note that -(x)^{4} is less than 0, and (-x)^{4} is greater than 0.

Now for negative exponents. **Here is a simple rule is the best way to explain it:**

x^{-n} = 1/x^{n}

A negative exponent indicates a value *reciprocal to *the value with a positive exponent. It’s good practice to “translate” any exponential expressions with negative exponents to their reciprocal positive forms. Seeing it both ways can help you make sense of problems.

5^{-4} = 1 / (5^{4}) = 1 / 625

17^{-2} = 1 / (17^{2}) = 1 / 289

(9 / 16)^{-2} = (16 / 9)^{2} = 256 / 81

Integer bases with negative exponents go under a numerator of 1; fractional bases with negative exponents simply flip. Let’s look at some “double negative” exponential expressions.

(-6)^{-3} = 1 / (-6)^{3} = 1 / -216

(-2)^{-10} = 1 / (2^{10}) = 1 / 1024

(-4 / 3)^{-4} = (-¾)^{4} = 81 / 256

Let’s get into some official GMAT problems. Be careful with this first one!

From the consecutive integers -10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?

- (-10)
^{20} - (-10)
^{10} - 0
- -(10)
^{19} - -(10)
^{20}

One incorrect answer is chosen far more often than any other on this problem: answer choice D. Trying to minimize the product, many people consider taking the maximum number (20) of the lowest value (-10). The common mistake is then thinking that -(10)^{20} (answer choice E) is actually a positive value since it involves an even number (20) of negative factors. Many people then take “the next best thing” in answer choice D, which shifts the exponent to the next odd number down from 20.

In fact, -(10)^{20} does not involve an even number of negative factors, since the negative sign is *excluded *from the exponential expression by the parentheses. Answer E means “take 10^{20} and make it negative.” It is true that taking 20 negative tens and multiplying them all together produces a large positive value (the opposite of what we are aiming for on this problem), but this misguided idea is notated by answer choice A – not answer choice E. Remember that the answer choices notate the *product *of the 20 factors, not necessarily a condensed list of the 20 factors.

It is possible to choose 20 integers from -10 to 10 inclusive that, when multiplied, yield a product of -(10)^{20} (answer choice E). The “least possible value” is obtained by finding the *greatest absolute value (distance from 0) in negative form*. So we want all 20 of our factors to be either 10 or -10 since this will maximize the absolute value (distance from 0) of the product. To ensure that the product is also negative, we simply need an odd number of negative tens. We can use nineteen negative 10s and 1 positive 10, 1 negative 10 and 19 positive 10s, or any odd combination in between. Any of these options will yield a product of -(10)^{20}. Read the notation carefully!

Let’s try another:

The value of 2^{(-14)}+2^{(-15)}+2^{(-16)}+2^{(-17)}/5 is how many time the value of 2^{(-17)}?

- 3/2
- 5/2
- 3
- 4
- 5

This problem benefits from the skill of noticing patterns and “checking” them. You should see the pattern in the numerator and generalize it by saying “the negative exponent on the 2 keeps decreasing by 1.” Then you can see how this pattern “works” by checking a single case.

2^{-17} = 1 / 2^{17}

2^{-16} = 1 / 2^{16}

Since 2^{17} = 2 * 2^{16}, (1 / 2^{17}) is half the value of (1 / 2^{16}). Or, to say it a more useful way, 2^{-16} = 2 * 2^{-17} This pattern will continue through the numerator. Since we are looking for how many copies of 2^{-17} we have in this expression, we can replace 2^{-17} with 1 and follow the pattern.

(2–^{14} + 2^{-15} + 2^{-16} + 2^{-17}) / 3

(8 + 4 + 2 + 1) / 3

15 / 3 = 3

**And the correct answer is C.**

Here’s a final problem for today:

*a *is a nonzero integer. Is

a^{2}greater than 1?

*a*< -1

*a*is even.

To evaluate statement 1, simply start by checking a = -2. (-2)^{-2} = 1 / (-2)^{2} = ¼. Moving on to a = -3, (-3)^{-3} = 1 / (-3)3 = 1 / -27. This time the value is negative, but the positive value from a = -2 is still less than 1. If you imagine continuing with a = -4, a = -5, etc., you will just keep making smaller and smaller fractions. **Statement 1 alone is sufficient.**

Statement 2 on its own is easy to check since we already know from checking statement 1 that some even values for the variable a yield an aa with a value less than 1. And it shouldn’t be hard to imagine an even value for variable a where aa is greater than 1. For example, 2^{2} = 4 and 4^{4} = 256. **So statement 2 alone is insufficient, and the correct answer is A.**

If you went with answer choice C, here’s what might have happened. Noticing that statement 1 tells you the base is negative, you might have seen next that statement 2 tells you the exponent is even. You might have thought that this “evenness” of the exponent makes the difference since it determines the positivity or negativity of the expression. Very often in DS problems with negative bases, the even/odd identity of the exponent really matters. But in this case, it’s a trap, because we were asked whether a^{a} is *greater than 1* (not 0), and the fact that the exponent is *also* negative means that it’s even/odd identity is irrelevant – the value is always less than 1.

The rules governing negative exponents and negative bases are simple, but the GMAT and EA problems that employ these rules can catch you if you aren’t careful. Next time we will look at another tricky exponential scenario: when the base is between -1 and 1.

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**Contributor: ***Elijah Mize (Apex GMAT Instructor)*