Negative and Zero Exponents: Additional Practice (Set 2)

Write $\tfrac{1}{8}$ as a power of $2$ using the negative-exponent rule.

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

So $2^n = 2^{-3}$, which gives $n = -3$.

Write $\tfrac{1}{81}$ as a power of $3$, since $81 = 3^4$.

$\frac{1}{81} = \frac{1}{3^4} = 3^{-4}$

So $3^n = 3^{-4}$, which gives $n = -4$.

Multiply the exponents; two negatives give a positive.

$\left(2^{-2}\right)^{-3} = 2^{(-2)(-3)} = 2^{6} = 64$

The base is $-3$, and the $-2$ exponent takes the reciprocal of $(-3)^2$. Square first: two negatives cancel.

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

The value is positive, because an even power of a negative base is positive and the reciprocal keeps the sign.

Use $7^0 = 1$ and $7^{-1} = \tfrac{1}{7}$, then subtract.

$1 - \frac{1}{7} = \frac{7}{7} - \frac{1}{7} = \frac{6}{7}$

Take the reciprocal of $(-2)^3$. An odd power keeps the base's negative sign.

$(-2)^{-3} = \frac{1}{(-2)^3} = \frac{1}{-8} = -\frac{1}{8}$

The result is negative because the base is negative and the exponent is odd, not because the exponent is negative.

Each is a unit fraction: $\tfrac{1}{3}, \tfrac{1}{4}, \tfrac{1}{2}, \tfrac{1}{5}$.

$\frac{1}{2} > \frac{1}{3} > \frac{1}{4} > \frac{1}{5}$

A smaller base gives a larger reciprocal, so $2^{-1} = \tfrac{1}{2}$ is the largest.

Write $0.01$ as a power of $10$.

$0.01 = \frac{1}{100} = \frac{1}{10^2} = 10^{-2}$

So $10^{n} = 10^{-2}$, which gives $n = -2$.

Write $16$ as a power of $\tfrac{1}{2}$. Since $\left(\tfrac{1}{2}\right)^{-1} = 2$,

$\left(\frac{1}{2}\right)^{-4} = 2^{4} = 16$

So $n = -4$.

Flip the fraction, then square it.

$\left(\frac{2}{5}\right)^{-2} = \left(\frac{5}{2}\right)^{2} = \frac{5^2}{2^2} = \frac{25}{4}$

Substitute $h = 4$ into $2^{-h}$.

$2^{-4} = \frac{1}{2^4} = \frac{1}{16}$

Halving four times shrinks the size to $\tfrac{1}{16}$, a positive fraction.

A negative exponent in the denominator climbs to the numerator, and $2^0 = 1$.

$\frac{1}{3^{-2}} = 3^{2} = 9, \qquad 9 + 1 = 10$