$0^n = 0 \qquad \{n \neq 0\}$

$a^0 = 1 \qquad \{a \in \mathbb{R}\}$

$a^m \times a^n = a^{m+n} \qquad \{m, n \in \mathbb{Q}\}$

$a^m \div a^n = a^{m-n} \qquad \{a \neq 0\}$

$(a^m)^n = a^{m \times n}$

$(ab)^m = a^mb^m$

$\sqrt[n]{a^m} = a^{\frac{m}{n}}$

$\sqrt[n]{a}\times\sqrt[n]{b} = \sqrt[n]{ab}$

$\displaystyle \frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}}$

$\displaystyle \sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}$

$a^{-1} = \frac{1}{a}$

$a^{-m} = \frac{1}{a^m}$

If $a^m = a^n$ then $m = n \qquad \{a \neq 0, a \neq \pm 1\}$

If $a^m = b^m$ then $m = 0 \qquad \{a \neq b\}$

Logarithms

If $a^n = b$ then $n = \log_a b \qquad \{a>0, a\neq 1, b>0\}$

Rather than remembering, you can derive the above from only two properties below. Start by applying $\log{x}$ to both sides of the equation \(a^n = b\).

$\log_a mn = log_a m + \log_a n$

$\log_a\frac{m}{n} = \log_a m - \log_a n$

$\log_a m^n = n\log_a m$

$\displaystyle \log_a b = \frac{\log_n b}{\log_n a}$

$\displaystyle \frac{1}{\log_b a} = \log_a b$

$\displaystyle \log_\frac{1}{a}b = -\log_a b$

$\log_a b \times \log_b a = 1$