What is $\frac{(-2)^{x}}{2^{x-1}}$

Question

The title says it all:

$$\frac{(-2)^{x}}{2^{x-1}}$$

How is this computed? I'm reviewing the finer points of exponents so a thorough explanation would be most appreciated!

Answer 1

Laws of exponents say that $(ab)^x=a^xb^x$, and $a^{x+y}=a^xa^y$. Therefore $$\frac{(-2)^x}{2^{x+1}}=\frac{(-1)^x2^x}{2\cdot 2^x}=\frac{(-1)^x}{2}.$$ This expression is a real number if $x$ is an integer. If $x$ is an even integer, it is equal to $1/2$. If $x$ is an odd integer, then it is equal to $-1/2$.

Answer 2

Hint: Note that $(-2)^x$ can be written as $((-1)(2))^x = (-1)^x(2)^x$.