Exponents: A new kind of super powerIn this lesson we'll be exploring whole number exponents and some rules for manipulating them and simplifying expressions involving them! Exponents, on a basic level, are shorthand notation for repeated multiplying, just like multiplication is in a sense shorthand for repeated addition. 

Just when you think things are about as confusing as they can get (I mean you made it through fractions, what more can they ask of you?), the evil masterminds behind mathematics start stacking numbers on numbers.
Let's look at the basic structure of a number raised to a positive integer "power". (We will address other kinds of powers in another lesson). \( m^n \) simply means "n factors of m all multiplied together". Of course the easiest way to understand is to give a few a try! 
RECALL: A positive integer is a whole number greater than zero, commonly referred to as the "counting" numbers, and technically termed the "natural numbers". TERMS: In the example on the left, "m" is called the base and "n" is called the exponent. 
\( 4^2 \) or "2 factors of 4 multiplied together"

So that would be \( 4 \times 4 = 16 \)!

\( 3^4 \) or "4 factors of 3 multiplied together"

So this time we'd have \( 3 \times 3 \times 3 \times 3 = 81 \)

IMPORTANT  Be careful with the use of parenthesis and negative base numbers, a mistake often and easily made by students is getting the following two types of problems mixed up (and they yield very different results!)
\( (2)^4 \) or "4 factors of 2 multiplied together"
\( 2^4 \) or "the negative OF (4 factors of 2 multiplied together)" 
That's \( 2 \times 2 \times 2 \times 2 = 16 \)
and THIS one is \( (2 \times 2 \times 2 \times 2) = 16 \) 
This difference is simply an application of the rules of "order of operations". Exponents override multiplication (which is what the outside negative is), but not containers like parenthesis.
What about exponents that aren't positive integers?In the opening part of the lesson I said we would be looking only at positive integer exponents. That is because some different things happen when we apply exponents that are negative and/or fractional and we will be investigating these in later lessons. The only thing left to be curious about is a zero exponent, and luckily, that's a really easy case to remember:
"any number raised to a 0 power is 1". 
\( 3^0 = 1 \ , \ 4^0 = 1 \ , \ 123,432^0 = 1 \)

How great is that?

Ok, so what now?So we know what the power means, and it's no longer a mysterious hat for another number. What if I had several of these critters in a row, or in LAYERS? Something like:
\[ (3^2+(5^2)^3)^7+(3^2)(3^8) \] Well it turns out that the required ideas aren't too crazy, let's check them out. 
Multiplying "Like Bases" 
This first rule can only be applied when we have two exponential expressions being multiplied that have the same base. Some examples of this are: \( (3^2)(3^3) \), \( (p^{10})( p^5) \), and \( (3+x)^2 (3+x)^3 \). It doesn't matter how crazy the bases are, as long as they are the same!

So when we have this situation we can apply the following rule that you must learn to cherish. Cherish it.
\[ (a^b)(a^c) = a^{b+c} \]
That is, when we multiply like bases, we add the exponents. Let's see it in action with the three examples above.
\[ (a^b)(a^c) = a^{b+c} \]
That is, when we multiply like bases, we add the exponents. Let's see it in action with the three examples above.
\[ (3^2)(3^3) = 3^{2+3} = 3^5 = 243 \]
Done! 
\[ (p^{10})(p^5)= p^{10+5} = p^{15} \]
Notice that we did not get a numerical answer for this, because we don't know what \( p \) is! Despite this, the rules for exponents still hold. This is vital in the upcoming lesson on Polynomials. 
\[ (3+x)^2(3+x)^3 = (3+x)^{2+3}=(3+x)^5 \]
Now, we could expand this further, but that lesson will come with the lesson on Polynomials. 
Let's consider for a moment why this works, because it's pretty straightforward. Remember from above that \(3^2 \) means \( 3 \times 3 \). Well if I have \( (3^2)(3^3) \), then I have \( (3 \times 3) \times ( 3 \times 3 \times 3) \), but that's just five copies of three, hence \( 3^5\)! This thinking helps with the next two rules as well.
Exponents over Multiplication 
This is a little bit of wacky working, but in short it is the rule that lets us handle something like \( (3x)^3 \) and \( (4ab)^2 \). Instead of having to write out so many repeated factors we can notice that since multiplication is commutative (order doesn't matter), then
\[ (3x)^3 = (3x)(3x)(3x) = (3)(3)(3)(x)(x)(x) = 3^3x^3! \] 
In general
\[ (ad)^c = a^cd^c \]
So what about \( (4ab)^2 \)? We get \( 4^2a^2b^2 = 16a^2b^2!\)
\[ (ad)^c = a^cd^c \]
So what about \( (4ab)^2 \)? We get \( 4^2a^2b^2 = 16a^2b^2!\)
AVOID "FRESHMAN'S DREAM" \((a+b)^2 = a^2 + b^2 \) THIS IS FALSE. THIS IS FALSE. THE RULE (EXPONENTS OVER MULTIPLICATION) ONLY WORKS OVER MULTIPLICATION! \( (a+b)^2 = a^2 +2ab + b^2 \) and will be covered in the lesson on polynomials! 

Raising "Powers to Powers" 
This next rule applies when we have an exponential expression being itself raised to a power. Some examples of this are
\[ (3^2)^3 , \ \ ((m+n)^6)^4, \ \ (((4^2)^2)^2)^2 \] 
Now let's think about this like we did at the end of the last section (Multiplying Like Bases). If I have \( (3^2)^3 \) then I have
\[ (3^2)^3 = (3^2) \times (3^2) \times (3^2) = (3 \times 3) \times ( 3 \times 3) \times (3 \times 3) = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6! \]
What's happening here? We are getting three copies OF two copies of the number 3...it might lead you to suspect then that
\[ (a^b)^c = a^{bc} \]
That is, when we raise powers to powers, we multiply the exponents. Let's see it work.
\[ (3^2)^3 = (3^2) \times (3^2) \times (3^2) = (3 \times 3) \times ( 3 \times 3) \times (3 \times 3) = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6! \]
What's happening here? We are getting three copies OF two copies of the number 3...it might lead you to suspect then that
\[ (a^b)^c = a^{bc} \]
That is, when we raise powers to powers, we multiply the exponents. Let's see it work.
\[ ((m+n)^6)^4 = (m+n)^{6 \cdot 4} = (m + n)^{24} \]

\[ (((4^2)^2)^2)^2 = ((4^2)^2)^{4} = (4^2)^8 = 4^{16} \]

Pass one!
\[4^2x^2+ 2^{3 \cdot 2}  (3x)^{2 \cdot 2} \] 
Pass two!
\[ 16x^2 + 2^6 (3x)^4 \] 
Pass three!
\[ 16x^2 + 64  3^4 \cdot x^4 \] 
Pass four!
\[ 16x^2 + 64  81x^4 \] Done! 