Negative and Fractional Exponents
A negative exponent means reciprocal; a fractional exponent means a root. Neither makes the number negative.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
$x^{-n} = \frac{1}{x^n}$, and $x^{1/n} = \sqrt[n]{x}$. A fractional exponent like $\frac{3}{4}$ combines a power and a root.
How to solve one, step by step
Example: evaluate $16^{3/4}$.
- Take the fourth root: $16^{1/4} = 2$.
- Cube it: $2^3 = 8$.
The mistakes that cost points
- Treating a negative exponent as a negative number. $x^{-2}$ is $\frac{1}{x^2}$, a positive reciprocal — not $-x^2$.
- Confusing x^(1/2) with x/2. A one-half exponent is a square root, not division by two.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
If $x^{-2} = \frac{1}{k}$ and $x = 4$, what is the value of $k$?
- A$8$
- B$\frac{1}{16}$
- C$16$
- D$-16$
Show solution
Answer: C, $16$. Substitute $x = 4$: $4^{-2} = \frac{1}{4^2} = \frac{1}{16}$. Since $x^{-2} = \frac{1}{k}$, we have $\frac{1}{16} = \frac{1}{k}$, so $k = 16$.
Medium
The equation $x^{-3} = \frac{1}{k}$ holds when $x = 2$. What is the value of $k$?
- A$\frac{1}{8}$
- B$8$
- C$-6$
- D$-8$
Show solution
Answer: B, $8$. Substitute $x = 2$: $2^{-3} = \frac{1}{2^3} = \frac{1}{8}$. Since $x^{-3} = \frac{1}{k}$, we get $\frac{1}{8} = \frac{1}{k}$, so $k = 8$.
Hard
If $\left(9x^2\right)^{\frac{1}{2}} = k - 1$ and $x = 4$, what is the value of $k$?
- A$73$
- B$13$
- C$-11$
- D$7$
Show solution
Answer: B, $13$. Substitute $x = 4$: $\left(9 \cdot 16\right)^{\frac{1}{2}} = 144^{\frac{1}{2}} = \sqrt{144} = 12$. So $12 = k - 1$, giving $k = 13$.
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