1. Simplifying Expressions with Integral Exponents
In this section we learn some important Laws of Exponents.
Back in the chapter on Numbers, we came across examples of very large numbers. (See Scientific Notation). One example was the mass of the Earth, which is about:
`6 × 10^24` kg
In this number, the `10` is raised to the power `24` (we could also say "the exponent of `10` is `24`").
The number `10` is called the base and `24` is called the exponent (or power).
Now, the number `10^24` means:
`10^24 = 10 ×10 ×10 ×10 ×10 × ... ×10` [We multiply `24` lots of `10`.]
`= 1,000,000,000,000,000,000,000,000` [That's `1` with `24` zeros]
Exponents give us a very convenient way of writing very large and very small numbers. They are also very handy for making algebra easier because it is more compact. Let's now give a general definition for any number (or any variable) raised to an "integral exponent":
Definition: am means "multiply m lots of a together"
am = a × a × a × a × a × ... × a
[There are m lots of a in our multiplication.]
Note 1: "Integral exponent" means the exponent is a whole number [That is, an integer]
Note 2: The above definition only really holds if m is a positive integer, since it doesn't make a lot of sense if m is negative. (You can't multiply something by itself negative 3 times! And what does multiplying something by itself 0 times mean?). In such cases we have to rely on patterns and conventions to define what is going on. See below for zero and negative exponents.
Example 1: Integral Exponents
(1) y5 = y × y × y × y × y [There are `5` lots of y being multiplied together.]
(2) `2^4= 2 × 2 × 2 × 2 = 16` [There are `4` lots of `2` being multiplied together.]
(3) `10^6= 10 × 10 × 10 × 10 × 10 × 10 = 1,000,000` [There are `6` lots of `10` being multiplied together.]
Multiplying Expressions with the Same Base
Let's start with an example. Once you get the hang of this, it makes writing math a whole lot easier.
Say we need to multiply 2 large numbers, `10^8` and `10^5`. Now, if we write it out in full, we would need to write:
`10^8 = 10 ×10 ×10 ×10 ×10 ×10 ×10 ×10` (`8` lots of `10` multiplied together)
`10^5 = 10 ×10 ×10 ×10 ×10` (`5` lots of `10` multiplied together)
`10^8 × 10^5 = (10 ×10 ×10 ×10 ×10 ×10 ×10 ×10) × (10 ×10 ×10 ×10 ×10)`
Now, if you count them all up, you will have `13` lots of `10` multiplied together.
So we can conclude that
`10^8 × 10^5 = 10^13`
This is very tedious and there must be an easier way. We could add the exponents when multiplying numbers with the same base. Let's see a general definition.
Definition: `a^m × a^n = a^(m+n)`
Let's see how this works with an example involving a variable, b:
`b^5 × b^3 = (b × b × b × b × b) × (b × b × b)`
Our final answer is equivalent to `b^(5+3)`.
Dividing Expressions with the Same Base
When we divide expressions with the same base, we need to subtract the exponent of the number we are dividing by from the exponent of the first number. In general, we can write is as follows.
Definition: Dividing algebraic expressions
(Of course, `a ≠ 0`, and `m` and `n` are integers.)
It may be easiest to see how this one works with an example.
We cancel 2 of the b's from the numerator (the top) and the two b's from the denominator (the bottom) of the fraction. The result is equivalent to `b^(7 − 2)`.
We could also write this problem as
`b^7 ÷ b^2 = b^(7 − 2) = b^5`
Repeated Multiplication of a Number Raised to a Power
Next we consider the case where we have a base raised to some exponent, then we raise that to some other exponent.
For example, we may start with `p^3` and need to raise it to the power `2`. How do we do that? We'll see the answer in a minute. First, let's look at a general definition.
`(a^m)^n = (a^m) × (a^m) × (a^m) × ... × (a^m) = (a^m)^n`
[We multiply n times]
So we write:
Definition: Repeated multiplication
`(a^m)^n = a^(mn)`
(Once again, we assume `a ≠ 0`, and `m` and `n` are integers.)
`(p^3)^2= p^3 × p^3` `= (p × p × p) × (p × p × p)= p^6`
A Product Raised to an Integral Power
In this section we have 2 numbers multiplied together, and we raise the result to some power. In this case, it has the same value as raising the first number to the power and multiplying by the second number raised to the power.
Definition: Product Raised to an Integral Power
`(ab)^n = a^nb^n`
`(5q)^3 = 5^3q^3 = 125q^3`
We have raised the `5` to the power `3` (giving us `125`) and we can't do anything else with `q^3`.
A Fraction Raised to an Integral Power
If we have a fraction raised to an integral power, we need to raise the top number to the power and divide by the bottom number raised to the power.
Definition: Fraction Raised to an Integral Power
(`n` is an integer.)
In this example, I have written out in full the meaning of `2/3` raised to the power `4`.
Answer: Raising the top and bottom numbers to the power of `5` gives:
Raising a Number to a Zero Exponent
Definition: `a^0 = 1` `(a ≠ 0)`
70 = 1
x0 = 1
(5a)0 = 1
Note 1: a0 = 1 is a convention, that is, we agree that raising any number to the power 0 is 1. We cannot multiply a number by itself zero times.
Note 2: In the case of zero raised to the power `0` (written `0^0`), mathematicians have been debating this for hundreds of years. It is most commonly regarded as having value 1, but is not so in all places where it occurs. That's why we write `a ≠ 0`.
Raising a Number to Negative Exponents
`a^(-n)=1/a^n` (Once again, `a ≠ 0`)
In this exponent rule, a cannot equal `0` because you cannot have `0` on the bottom of a fraction.
Explanation: 0 and Negative Exponents
Observe the following decreasing pattern:
34 = 81
33 = 27
32 = 9
31 = 3
For each step, we are dividing by `3`. Now, continuing beyond `3^1` and dividing by `3` each times gives us:
Summary - Laws of Exponents
`a^-n=1/a^n\ (ane0) `
[Note: These laws also apply in the next section, Fractional Exponents.]
Let's now try some mixed examples where we have integral exponents.
(a) Simplify `a^5 × a^-3`
(b) Simplify `a^3 × a^-5`
Simplify `(2^3 × 2 ^-4)^2`
The importance of brackets
Note the following differences carefully:
(-5x)0 = 1, but -5x0 = −5.
In the first one, we are raising everything in brackets to the power `0`, so the answer is `1`.
In the second one, we are only raising the x to the power `0`, then we are left with `−5 × 1 = −5`
(−5)0 = 1, but −50 = −1.
Simplify (2a + b-1)-2
Question (1) Simplify: `(5an^-2)^-1`
Question (2) Simplify: `((a^-2)/(b^2))^-3(a^-3/b^5)^2`
Question (3) Simplify: `(2a - b^-2)^-1 `