# Formula for exponents?

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.

• You're going to have to be more specific, I don't think in this generality there's a meaningful answer to your question. – Gregory Grant May 13 '15 at 20:31
• This is a genuine question, not trying to be hurtful: How can I be more specific? I need a formula to generate a^b that works universally (Of course the formula can't be a^b) – user240331 May 13 '15 at 20:32
• A formula for exponents in what context? What is $a$? What is $b$? Why is $a^b$ not simplified enough as it is? – JMoravitz May 13 '15 at 20:33
• Can you show us an example of a "formula" that works for you if $a$ and $b$ are both positive? Just so we can have a better idea of what you mean. – Gregory Grant May 13 '15 at 20:36
• @user240331 that particular formula is a direct consequence of the property that $b^1=b$ and that $b^{n+m} = b^n\cdot b^m$. The result that you write though is only valid when $n$ is a natural number like 1,2,3,4,5,... (after all, what does it mean to multiply something four-and-a-half times?) – JMoravitz May 13 '15 at 20:56

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

• Just pointing out, $x^n\neq \frac{1}{x^n}$ unless $x^n = 1$. Also, you can use & to offset things in cases. I'm taking the liberty of editing the mistakes and odd typesetting. Furthermore, you had "$x>0$" instead of $n>0$. Your third case seemed out of place as well. $x^n=\frac{1}{x^{-n}}$ is always true, not just when $n<0$. – JMoravitz May 14 '15 at 2:28

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.

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.

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: $2^{5}=32$

• May be this answer is better for an unclear understanding. – Sufyan Naeem May 13 '15 at 20:48