Any non-zero number raised to the power $0$ equals $1$. This isn't a special rule to memorise separately — it falls straight out of the quotient law. Since $\dfrac{a^n}{a^n}=1$ for any non-zero $a$, and the quotient law says $\dfrac{a^n}{a^n}=a^{n-n}=a^0$, it must be true that $a^0=1$.

A negative index does not make a number negative — it means "reciprocal." $a^{-n}$ is simply $1$ divided by $a^n$:

The denominator of a fractional index tells you which root to take; the numerator tells you which power to raise it to (in either order — the result is the same):

Treat the two parts separately: evaluate the fractional index first to get a root-and-power result, then flip that result into a reciprocal because of the negative sign. For $16^{-3/4}$:

1. Deal with the fractional index: $16^{3/4} = \left(\sqrt[4]{16}\right)^3 = 2^3 = 8$.
2. Apply the negative sign as a reciprocal: $16^{-3/4} = \dfrac{1}{8}$.

Apply the index to every factor inside a bracket separately, using the standard laws $a^m \times a^n = a^{m+n}$, $a^m \div a^n = a^{m-n}$, and $(a^m)^n = a^{mn}$ at each stage.

First deal with the number: $27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$, so $27^{-2/3} = \dfrac{1}{9}$.

Then deal with the variable using power-of-a-power: $\left(x^6\right)^{-2/3} = x^{6\times(-2/3)} = x^{-4}$.

Combining and rewriting with a positive index: $\dfrac{1}{9}x^{-4} = \dfrac{1}{9x^4}$.

Rewrite both sides of the equation using the same base, then simply equate the two exponents and solve the resulting (usually linear) equation.

Worked Example

Solve for $x$: $\quad 4^{3x-1} = 8^{x+2}$

1. Write both sides using base $2$: $4 = 2^2$ and $8=2^3$.

1. Apply the power-of-a-power rule to both sides:

1. Since the bases now match, equate the exponents:

1. Solve the linear equation: