Negative and Zero Exponents: Practice

Evaluate each term, then add over a common denominator.

$2^{-3} = \frac{1}{8}, \qquad 2^{-1} = \frac{1}{2} = \frac{4}{8}$

So the sum is

$\frac{1}{8} + \frac{4}{8} = \frac{5}{8}$

Add the exponents on top, then subtract the bottom exponent.

$a^{3} \cdot a^{-5} = a^{3 + (-5)} = a^{-2}$

Then divide:

$\frac{a^{-2}}{a^{-4}} = a^{-2 - (-4)} = a^{2}$

Flip the fraction, then square it.

$\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^{2} = \frac{3^2}{2^2} = \frac{9}{4}$

Add inside the parentheses first, over a common denominator.

$3^{-1} + 6^{-1} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{1}{2}$

Then the outer $-1$ exponent inverts the result:

$\left(\frac{1}{2}\right)^{-1} = 2$

The base is $-2$ and the $-2$ exponent calls for the reciprocal of $(-2)^2$. Square first, where two negatives cancel.

$(-2)^{-2} = \frac{1}{(-2)^2} = \frac{1}{4}$

The value is positive: a negative exponent never makes the result negative, and an even power of a negative base is positive.

Add the exponents with the product rule.

$3^{-2} \cdot 3^{4} = 3^{-2 + 4} = 3^{2} = 9$

Evaluate each value.

$2^{-2} = \frac{1}{4}, \quad 2^{-1} = \frac{1}{2}, \quad 2^{0} = 1$

From smallest to largest, $\tfrac{1}{4} < \tfrac{1}{2} < 1$, so $2^{-2} < 2^{-1} < 2^{0}$.

Rewrite each power with a positive exponent, then divide.

$4^{-1} = \frac{1}{4}, \qquad 2^{-3} = \frac{1}{8}$

Dividing by a fraction multiplies by its reciprocal:

$\frac{1/4}{1/8} = \frac{1}{4} \times 8 = 2$

Divide the coefficients, then subtract exponents for each base.

$\frac{6}{2} = 3, \quad x^{-2-1} = x^{-3}, \quad y^{1 - (-3)} = y^{4}$

Moving the negative power of $x$ to the denominator gives

$\frac{3 y^4}{x^3}$

Simplify inside the parentheses first, then invert with the $-1$ exponent.

$\frac{x^{2}}{x^{5}} = x^{2-5} = x^{-3}$

Then

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

Rewrite each term and add over a common denominator.

$2^{-1} + 3^{-1} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$

Check each value.

$3^{-2} = \frac{1}{9}, \quad \frac{1}{3^2} = \frac{1}{9}, \quad \left(\frac{1}{3}\right)^2 = \frac{1}{9}$

The first option has a negative exponent, which flips the fraction:

$\left(\frac{1}{3}\right)^{-2} = 3^2 = 9$

So $\left(\tfrac{1}{3}\right)^{-2} = 9$ is the one that does not equal $\tfrac{1}{9}$.