10 Pitfalls to Avoid When Working with Exponents
In algebra, the rules used when working with exponents are straightforward and consistent. Challenges arise, though, when applying the rules or knowing how to apply the rules in situations where the problem is more complicated and doesn’t look exactly like the rule.
Raising to a power
The rules for raising a power to a power or two factors to a power are
Basically, these rules say that you multiply the original exponent times the power. This looks fine in this format, but here are some common errors:

(a^{3})^{5} ≠ a^{8}, where the exponents are added instead of multiplied. This should be (a^{3})^{5} = a^{15}.

(2x^{3}y^{4})^{5} ≠ 2x^{15}y^{20}, where the coefficient is forgotten. This should be (2x^{3}y^{4})^{5} = 2^{5}x^{15}y^{20} = 32x^{15}y^{20}.
Negative exponents
The rules for dealing with negative exponents include
The last rule is just a special case of the first rule listed. It’s here for emphasis.
The negative exponents were created for ease in combining factors with the same base. But some abuse often occurs, such as the following:
Here, the coefficient hasn’t been assigned a negative exponent. This should read as follows:
Or, you could just leave the 6 in the denominator and write
Powers of roots
When changing from a radical expression to one using fractional exponents, the rules are
A root is indicated with a fractional exponent. The root always goes in the denominator of the fraction. When a power of the root is involved, place it in the numerator of the fraction.
A common error is the following:
This places the root (fourth root) in the numerator, not the denominator. Instead, it should be written
Another challenge occurs when going from the fractional root to the radical. When rewriting a^{5/3}, you use the following:
Or you can write the power outside the radical as follows:
Forgetting resulting factors
Factoring expressions is a basic process in mathematics. Taking out a greatest common factor (GCF) is usually the first choice you make when performing a factorization. A problem arises when a division result isn’t indicated:
You have to indicate the result of each division:
Factoring fractional exponents
Performing factorizations involving fractional exponents — especially negative fractional exponents — can be sticky. For example, when factoring 4a^{1/2} – 3a^{–1/2} you first have to decide on what the GCF is. The rule with powers of the same variable is to divide out the lower of the two powers. In this case, the lower power is
And the rule when dividing terms with the same base is
In this case, 4a^{1/2} – 3a^{–1/2} = a^{–1/2}(4a–3). Remember, when dividing, you subtract exponents, and
Hidden exponents
In mathematics, there are many conventions used when writing expressions. For example, when you write the number 6, you assume that it’s +6 and that the power on the 6 is a 1 and that there’s a decimal point to the right of the 6. It would take much longer to write numbers if each of those symbols had to be written in. The challenge is to not forget that the notations are there.
The hidden exponents can get lost when factoring fractional expressions. For example:
First, the rule is that you have to divide each term in the fraction by the same value. Second, the GCF of the three terms is not a^{2}. The hidden exponent is on the 1 — because you can write the 1 as a^{0}, making the lowest power, or GCF, the term a^{0}. So, the actual factorization of this fraction is to leave it as is — dividing by a^{0} = 1 doesn’t change a thing.
Multiple negative exponents
Negative exponents are so handy, but they can also be problematic to the unprepared or those in a hurry. For example:
You say, “Oh, no, I’d never do that.” That’s good to hear, but don’t get caught in a quick fix with similar expressions. The proper way to deal with the expression is
Distributing over fractional exponents
Those fractional exponents keep coming up as problem children. You wouldn’t think that they’d be all that much trouble, especially because most people have been working with adding fractions since early grade school. It’s just that, when fractions are put into an exponential situation, sometimes those rules get forgotten. The rule I’m referring to here has to do with multiplying terms with the same base:
a^{x} a^{y} = a^{x}^{+}^{y}^{}
Applying this to a distribution, a common error is to multiply, rather than add:
a^{2}(a^{1/2} + a^{3/2}) ≠ a^{2} + a^{3}
Yes, it’s awfully tempting to eliminate those pesky fractional exponents by multiplying by 2, but the rule is to add the exponents. Here’s how it’s done:
a^{2}(a^{1/2} + a^{3/2}) = a^{2} a^{1/2} + a^{2} a^{3/2} = a^{2+1/2} + a^{2+3/2} = a^{5}^{/2} + a^{7}^{/2}^{}
Order of operations
According to the order of operations, you perform all powers and roots before multiplication and division. You perform multiplication and division before addition and subtraction. Of course, those grouping symbols can interrupt the process by requiring that you handle what’s in the grouping symbol, first. A really tempting move is to do the following:
2(a – 1)^{5} ≠ (2a – 2)^{5}^{}
This common error often occurs in situations where you have to evaluate an expression for some particular value of the variable. But, if you want to rewrite the expression without parentheses, you have to do the following:
The binomial theorem comes into play when raising binomials such as (a – 1) to a power.
Powering up binomials
The binomial theorem provides a way of determining the coefficients of the power of a binomial. The order of operations and rules of exponents are important here, because the following are common errors when performing those powers:
(a + b)^{2} ≠ a^{2} + b^{2}
(a + b)^{3} ≠ a^{3} + b^{3}
(a + b)^{4} ≠ a^{4} + b^{4}
When raising a binomial to a power, you’re actually multiplying that binomial the number of times indicated by the power:
(a + b)^{2} = (a + b)(a + b)
(a + b)^{3} = (a + b)(a + b)(a + b)
(a + b)^{4} = (a + b)(a + b)(a + b)(a + b)
You then use the binomial theorem or Pascal’s triangle to help you fill in the correct exponents and coefficients:
(a + b)^{2} = a^{2} + 2ab + b^{2}^{}
(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}
(a + b)^{4} = a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4}