~~DISCUSSION|Math Discussion~~

<latex>a^na^m = a^{n+m}</latex>

<latex>\frac{a^m}{a^n} = a^{m-n}</latex>

<latex>a^{-n} = \frac{1}{a^n} = (\frac{1}{a})^n</latex>

<latex>a^0 = 1, a \neq 0</latex>

<latex>(ab)^m = a^mb^m</latex>

<latex>(a^m)^n = a^{mn}</latex>

Take a perfect square number, let's use nine. Multiply the number by itself ( nine times is 81 ); this is written <latex>9*9=81</latex>. 9 times 9 can also be written as <latex>9^2=81</latex>, that is, 9 squared, or 9 raised to the second power. Since the substitution property allows 9 to be to be written as <latex>3*3=9</latex>, the previous could be written as <latex>(3*3)^2=81</latex>. That is, the product of 3 repeated 3 times, repeated another 9 (or <latex>3*3</latex>) times is 81. 3 times 3 can be written as <latex>3^2</latex>, therefore, <latex>(3^2)^2=81</latex>. Written out long-hand, the product of 3 squared raised to the second power looks like this: <latex>(3*3)*(3*3)=81 \to (3*3)^2 \to 3^{2*2} \to 3^4</latex>.

• <latex>9*9=81 \to 9^2=81 \to (3*3)^2=81 & \to (3^2)^2=81 \to 3^{2*2} = 81 \to 3^4 = 81</latex>

<latex>(\frac{a}{b})^m = \frac{a^m}{b^m}</latex>

<latex>|a^2| = |a|^2 = a^2</latex>

<latex>\sqrt{a} = a^\frac{1}{2}</latex>

<latex>\sqrt[n]{a} = a^\frac{1}{n}</latex>