As I've been helping people with algebra recently I've noticed, most of the people have forgotten their exponent rules, and instead of telling people what they are constantly, I thought I'd put some rules up to help you :)

Question

anonymous · Answer

$$\Huge \text{EXPONENTS}$$

$$a^1=a~~~8^1=8$$

$$a^0=1;~ a \neq 0 ~~~ \pi ^0 = 1$$

$$\frac{ a^m }{ a^n } = a ^{m-n}~~~\frac{ 81 }{ 9 }=\frac{ 3^4 }{ 3^2 }=3^{4-2}=3^2=9$$

$$a^m \times a^n = a ^{m+n}~~~ x^4 \times x^2 = x ^{4+2}=x^6$$

$$(a^m)^{n} ~~~(7^2)^{3} = 7^{2 \times3}=7^6=117,649$$

$$a ^{\frac{ m }{ n }} = \sqrt[n]{a^m}~~~27^{\frac{ 2 }{ 3 }} = \sqrt[3]{27^2}=9$$

$$a ^{-n}=\frac{ 1 }{ a^n }~~~6^{-2}=\frac{ 1 }{ 6^2 }=\frac{ 1 }{ 36 }$$ (this is the one people have the most trouble with, hopefully this helps)

$$(a \times b)^n=a^n \times b^n~~~(5 \times x)^3 = 5^3 \times x^3=125x^3$$

$$\left( \frac{ a }{ b } \right)^n=\frac{ a^n }{ b^n }~~\left( \frac{ 1 }{ 2 } \right)^3=\frac{ 1^3 }{ 2^3 }=\frac{ 1 }{ 8 }$$

$$\left( \frac{ a }{ b } \right)^{-n}=\left( \frac{ b }{ a }\right)^n=\frac{ b^n }{ a^n }~~~~~~~\left( \frac{ 2 }{ 3 } \right)^{-2}=\left( \frac{ 3 }{ 2 } \right)^2=\frac{ 3^2 }{ 2^2 }=\frac{ 9 }{ 4 }$$

sleepyjess · Answer

If anything to the power of 1 is itself, and anything to the power of zero is 1, what is $0^0$?

zarkon · Answer

$0^0$ is typically undefined

Indeterminate form is usually reserved for limits

nincompoop · Answer

you can say that 0^0 = 1 
argue it from 0!

anonymous · Answer

Thus
$$\ln 1= 0*\ln 0$$
$$0=\infty*0$$

anonymous · Answer

I meant - infinity**

misty1212 · Answer

ahh this old argument
don't buy in to it, let it be

anonymous · Answer

$$
\lim_{x \to 0}x^0 = 1\quad\text{and}\quad \lim_{x\to 0}0^x=0
$$This is why $0^0$ is an indeterminate form.  It can't be defined in a way to make these functions continuous.