Exponents are shorthand for repeated multiplication of the same thing by itself. For instance, the shorthand for multiplying three copies of the number 5 is shown on the right-hand side of the "equals" sign in (5)(5)(5) = 5^{3}. The "exponent", being 3 in this example, stands for however many times the value is being multiplied. The thing that's being multiplied, being 5 in this example, is called the "base".
This process of using exponents is called "raising to a power", where the exponent is the "power". The expression "5^{3}" is pronounced as "five, raised to the third power" or "five to the third".
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There are two specially-named powers: "to the second power" is generally pronounced as "squared", and "to the third power" is generally pronounced as "cubed". So "5^{3}" is commonly pronounced as "five cubed".
When we deal with numbers, we usually just simplify; we'd rather deal with "27" than with "3^{3}". But with variables, we need the exponents, because we'd rather deal with "x^{6}" than with "xxxxxx".
Exponents have a few rules that we can use for simplifying expressions.
To simplify this, I can think in terms of what those exponents mean. "To the third" means "multiplying three copies" and "to the fourth" means "multiplying four copies". Using this fact, I can "expand" the two factors, and then work backwards to the simplified form. First, I expand:
(x^{3})(x^{4}) = (xxx)(xxxx)
Now I can remove the parentheses and put all the factors together:
(xxx)(xxxx) = xxxxxxx
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This is seven copies of the variable. "Multiplying seven copies" means "to the seventh power", so this can be restated as:
xxxxxxx = x^{7}
Putting it all together, the steps are as follows:
(x^{3})(x^{4}) = (xxx)(xxxx)
= xxxxxxx
= x^{7}
Then the simplified form of (x^{3})(x^{4}) is:
x^{7}
Note that x^{7} also equals x^{(3+4)}. This demonstrates the first basic exponent rule:
Whenever you multiply two terms with the same base, you can add the exponents:
( x ^{m} ) ( x ^{n} ) = x^{( m + n )}
However, we can NOT simplify (x^{4})(y^{3}), because the bases are different: (x^{4})(y^{3}) = xxxxyyy = (x^{4})(y^{3}). Nothing combines.
Now that I know the rule (namely, that I can add the powers on the same base), I can start by moving the bases around to get all the same bases next to each other:
(a^{5} b^{3}) (a b^{7}) = (a^{5}) (a) (b^{3}) (b^{7})
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Now I want to add the powers on the a's and the b's. However, the second a doesn't seem to have a power. What do I add for this term?
Anything that has no power on it, in a technical sense, being "raised to the power 1". Anything to the power 1 is just itself, since it's "multiplying one copy" of itself. So the expression above can be rewritten as:
(a^{5}) (a) (b^{3}) (b^{7}) = (a^{5}) (a^{1}) (b^{3}) (b^{7})
Now I can combine:
(a^{5}) (a^{1}) (b^{3}) (b^{7}) = a^{5+1} b^{3+7} = a^{6} b^{10}
Putting it all together, my hand-in work would look like this:
(a^{5} b^{3}) (a b^{7}) = (a^{5} a^{1}) (b^{3} b^{7}) =
a^{6} b^{10}
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In the following example, there are two powers, with one power being "inside" the other, in a sense.
To do the simplification, I can start by thinking in terms of what the exponents mean. The "to the fourth" on the outside means that I'm multiplying four copies of whatever base is inside the parentheses. In this case, the base of the fourth power is x^{2}. Multiplying four copies of this base gives me:
(x^{2})^{4} = (x^{2})(x^{2})(x^{2})(x^{2})
Each factor in the above expansion is "multiplying two copies" of the variable. This expands as:
(x^{2})(x^{2})(x^{2})(x^{2}) = (xx)(xx)(xx)(xx)
Removing the parentheses, I get:
(xx)(xx)(xx)(xx) = xxxxxxxx
This is a string of eight copies of the variable. "Multiplying eight copies" means "to the eighth power", so this means:
xxxxxxxx = x^{8}
Putting it all together:
(x^{2})^{4} = (x^{2})(x^{2})(x^{2})(x^{2})
= (xx)(xx)(xx)(xx)
= xxxxxxxx
= x^{8}
Note that (x^{2})^{4} = x^{8}, and that 2 × 4 = 8. This demonstrates the second exponent rule:
Whenever you have an exponent expression that is raised to a power, you can simplify by multiplying the outer power on the inner power:
( x^{m} ) ^{n} = x ^{m n}
If you have a product inside parentheses, and a power on the parentheses, then the power goes on each element inside. For instance:
(xy^{2})^{3} = (xy^{2})(xy^{2})(xy^{2})
= (xxx)(y^{2}y^{2}y^{2})
= (xxx)(yyyyyy)
= x^{3}y^{6}
= (x)^{3}(y^{2})^{3}
Another example would be:
Warning: This rule does NOT work if you have a sum or difference within the parentheses. Exponents, unlike mulitiplication, do NOT "distribute" over addition.
For instance, given (3 + 4)^{2}, do NOT succumb to the temptation to say, "Hey, this equals 3^{2} + 4^{2} = 9 + 16 = 25", because this is wrong. Actually, (3 + 4)^{2} = (7)^{2} = 49, not 25.
When in doubt, write out the expression according to the definition of the power. For instance, given (x – 2)^{2}, don't try to do this in your head. Instead, write it out; "squared" means "multiplying two copies of", so:
(x – 2)^{2} = (x – 2)(x – 2)
= x(x – 2) – 2(x – 2)
= xx – 2x – 2x + 4
= x^{2} – 4x + 4.
The mistake of erroneously trying to "distribute" the exponent is most often made when the student is trying to do everything in his head, instead of showing his work. Do things neatly, and you won't be as likely to make this mistake.
Now that I know the rule about powers on powers, I can take the 4 through onto each of the factors inside. (I'll need to remember that, with the c, inside the parentheses it's "to the power 1".)
(a^{2})^{4} (b^{3})^{4} (c^{1})^{4}
= (a^{2×4}) (b^{3×4}) (c^{1×4})
= a^{8} b^{12} c^{4}
There is one other rule that may or may not be covered in your class at this stage:
Anything to the power zero is just "1".
This rule is explained on the next page. In practice, though, this rule means that some exercises may be a lot easier than they may at first appear:
Who cares about that stuff inside the square brackets? I sure don't, because the zero power on the outside means that the value of the entire thing is just 1. Ha!
[(3x^{4}y^{7}z^{12})^{5} (–5x^{9}y^{3}z^{4})^{2}]^{0} = 1
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By the way, as soon as your class does cover "to the zero power", you should expect an exercise like the one above on the next test. It's a common trick question, designed to make you waste a lot of your limited time — but it only works if you're not paying attention.
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