A negative index is an instruction to find the reciprocal — it never makes the base negative. $a^{-n} = \dfrac{1}{a^n}$. So $3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}$, not $-9$. For a fraction raised to a negative power, flip the fraction and make the power positive: $\left(\dfrac{4}{5}\right)^{-2} = \left(\dfrac{5}{4}\right)^2 = \dfrac{25}{16}$.

For $a^{\frac{m}{n}}$, the denominator is the root and the numerator is the power: $a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m$. Without a calculator, always take the root first — the numbers stay small. $81^{\frac{3}{4}}$: 4th root of 81 is 3, then cube it: $3^3 = 27$. (Cubing 81 first and then rooting the huge result is needlessly hard.)

• $100^{\frac12} = \sqrt{100} = 10$
• $125^{\frac23}$: cube root of 125 is 5, then square: $5^2 = 25$

Deal with the parts in a fixed order: negative first (reciprocal), then root, then power.

Evaluate $16^{-\frac{3}{4}}$: Reciprocal: $\dfrac{1}{16^{\frac34}}$. 4th root of 16 is 2. Cube it: $2^3=8$. Answer: $\dfrac{1}{8}$.

If the bases are powers of the same prime, rewrite everything with that prime so the index laws ($a^m \times a^n = a^{m+n}$, $(a^m)^n = a^{mn}$) can be applied directly.

Simplify $9^5 \times 3^{-4}$: $9 = 3^2$, so $(3^2)^5 \times 3^{-4} = 3^{10} \times 3^{-4} = 3^{6}$.

Simplify $(8x^9)^{-\frac23}$: apply the power to each factor: $8^{-\frac23} \times x^{9 \times(-\frac23)}$. Now $8^{\frac23} = (\sqrt[3]{8})^2 = 4$, so $8^{-\frac23}=\frac14$, and $x^{-6}$. Combined: $\dfrac{1}{4x^6}$.

The same two-step order applies, just starting by flipping the fraction instead of a whole number.

Evaluate $\left(\dfrac{25}{16}\right)^{-\frac32}$: flip first: $\left(\dfrac{16}{25}\right)^{\frac32}$. Take the root (denominator 2): $\sqrt{\dfrac{16}{25}}=\dfrac45$. Then the power (numerator 3): $\left(\dfrac45\right)^3 = \dfrac{64}{125}$.

Rewrite both sides with the same base; once the bases match, the indices must be equal.

Solve $4^{-n} = 32$: both are powers of 2: $(2^2)^{-n} = 2^5 \Rightarrow 2^{-2n}=2^5 \Rightarrow -2n = 5 \Rightarrow n = -\dfrac52$.

Solve $4^n = 8$: $(2^2)^n=2^3 \Rightarrow 2n=3 \Rightarrow n=\dfrac32$ — a reminder that $n$ doesn't have to come out as a whole number.

Given that $3^{-n} = 0.2$, find the value of $(3^n)^2$.

1. $0.2 = \dfrac15$, so $3^{-n} = \dfrac15 = \dfrac{1}{3^n}$.
2. Comparing, $3^n = 5$.
3. $(3^n)^2 = 5^2 = 25$.

This "work backwards from a negative index" style is a common Grade 7–8 differentiator — the trap is trying to find $n$ itself rather than spotting that only $3^n$ is actually needed.