Formula for exponents?

Question

Is there a formula for exponents that works with both negative and positive powers? I have tried searching online but only found: If positive do this, if negative do this. Thanks.

EDIT: Ah, I see why I wasnt clear enough. I need a formula for exponents but it only needs to work with positive and negative integers.

Answer 1

$$ \large x^n= \begin{cases} \underbrace{x\cdot x \cdots x \cdot x}_{n \text{ times}}& n>0\\ 1 &n =0\\ \frac{1}{\underbrace{x\cdot x \cdots x \cdot x}_{|n| \text{ times}}}&n<0 \end{cases} $$

Answer 2

We could use the power series for the exponential function. In particular, we have $$ e^x = \sum_{k=0}^\infty \frac{x^k}{k!} $$ It would therefore follow that (for $a > 0$), $$ a^{x} = e^{\ln(a)x} = \sum_{k=0}^\infty \frac{[\ln(a)]^k x^k}{k!} $$ For other values of $a$, the value $a^x$ is not guaranteed to "make sense" in the context of the real numbers.

Answer 3

Exponentiation in general follows the following rules:

For all nonzero $x$

• $x=x^1$
• $x^a\cdot x^b = x^{a+b}$
• $x^0 = 1$

From these basic properties, we can see several things:

• $x^{-1}\cdot x = x^{-1}\cdot x^1 = x^0 = 1$ so $x^{-1}$ is in fact the multiplicative inverse of $x$
• $x^n = x\cdot x^{n-1}$
• $x^n = \underbrace{x\cdot x\cdots x}_{n~x\text{'s}}$ for a natural number $n$
• $x^{-1}=\frac{1}{x}$ since multiplicative inverses are unique.
• $x^{-a} = \frac{1}{x^a}$
• $(x^a)^b = x^{a\cdot b}$

For defining the exponentiation function for $x^n$ where $n$ is an integer, we can use the third result I listed above to relate it to multiplication and the fifth result if it were negative to simplify it first. For using exponents that are not integers, it requires a more careful definition, such as what Omnomnomnom provided in his answer.

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A short example that uses several of these properties:

Simplify the following expression: $$\frac{2^3\cdot 3^1\cdot 2^1}{2^{-1}\cdot 3^1}$$

Answer: $2^{5}=32$