Exponent Combination Laws

Theorem
Let $$a, b \in \R_+$$ be positive real numbers.

Let $$x, y \in \R$$ be real numbers.

Let $$a^x$$ be defined as $a$ to the power of $x$.

Then:
 * $$a^{x + y} = a^x a^y$$;


 * $$\left({a b}\right)^x = a^x b^x$$;


 * $$a^{-x} = \frac 1 {a^x}$$;


 * $$\left({a^x}\right)^y = a^{xy}$$;


 * $$\frac{a^x}{a^y} = a^{x-y}$$;


 * $$\left(\frac{a}{b}\right) ^x = \frac{a^x}{b^x}$$;


 * $$\left(\frac{a}{b}\right) ^{-x} = \left(\frac{b}{a}\right) ^x$$.

Proof
We use the definition $$a^x = \exp \left({x \ln a}\right)$$ throughout.


 * $$a^{x + y} = a^x a^y$$:

$$ $$ $$ $$


 * $$\left({a b}\right)^x = a^x b^x$$:

$$ $$ $$ $$


 * $$a^{-x} = \frac 1 {a^x}$$:

$$ $$ $$ $$


 * $$\left({a^x}\right)^y = a^{xy}$$:

$$ $$ $$


 * $$\frac{a^x}{a^y} = a^{x-y}$$:

$$ $$ $$


 * $$\left(\frac{a}{b}\right) ^x = \frac{a^x}{b^x}$$;

$$ $$ $$ $$


 * $$\left(\frac{a}{b}\right) ^{-x} = \left(\frac{b}{a}\right) ^x$$.

$$ $$