1.4 Powers and Radicals
- Page ID
- 152869
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- compute and simplify expressions with powers in them
- compute and simplify expressions with radicals in them, and relate them to powers
- deal with negative numbers in these contexts
Working With Powers (Exponents)
Powers (also called exponents) are when we write a lil guy upstairs to indicate that we're multiplying a number by itself multiple times. They look like this:
\[ 10^2 = 10 \cdot 10 = 100, \quad 2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32, \quad 5^3 = 5 \cdot 5 \cdot 5 = 125, \notag \]
In words, we say \( 10^2 \) as "ten squared" or "ten to the power of two" or if you're lazy like me, maybe "ten to the two." You'll hear \(5^3\) said as "five cubed," but for numbers bigger than 3, we just say "to the power of" whatever. The downstairs number is called the base, and the upstairs number is called the exponent. In all those examples (for this "repeated multiplication" interpretation of a power) the exponent was a positive integer. It's also possible to have negative powers, as well as fractional powers. In fact, any real number can be used as a power, even irrational numbers, but we won't see those much at this point. First, let's explore powers using a nice security blanket plain 'ol number. What's the most comforting number? Two? I agree. (Plus if you're in computer science, you have to memorize these anyway.) Let's raise him to some powers.
| 2 to the power of... | Result |
| 1 | \( 2^1 = 2 \) |
| 2 | \( 2^2 = 2 \cdot 2 = 4 \) |
| 3 | \( 2^3 = 2 \cdot 2 \cdot 2 = 8 \) |
| 4 | \( 2^4 = 2 \cdot 2 \cdot 2 \cdot 2 = 16 \) |
| 5 | \( 2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32 \) |
| \(k\), a positive integer | \( 2^k = 2 \cdot 2 \cdot ... \cdot 2 \) (\(k\) times) |
Okay, not too bad. I noticed something interesting up there while I was calculating the results. I didn't have to do all the multiplications every time, I could just look at the previous answer and multiply that by 2 again. So I observe that \(2^k = 2^{k-1}\cdot 2 \). Cool. What happens if I multiply two powers of 2 together?
\[ 2^2 \cdot 2^3 = (2 \cdot 2) \cdot ( 2 \cdot 2 \cdot 2 ) = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^5 \notag \]
Since it doesn't matter what order I multiply numbers in, I didn't need those parentheses at all! What do you observe? The new exponent turned out to be 5, which just so happens to be 2+3.
What happens if I divide two powers of 2?
\[ \frac{2^3}{2^2} = \frac{2 \cdot 2 \cdot 2}{2 \cdot 2} = \textcolor{red}{\frac{2 \cdot 2}{2\cdot 2} }\cdot \textcolor{green}{\frac{2}{1}} = \textcolor{red}{1} \cdot \textcolor{green}{2} = 2 \notag \]
What's the new power? Remember that \(2 = 2^1 \) and notice that \(1 = 3-2\)... Iiiinterestingggggg...
Finally, let's take a power of 2 and raise that whole thing to a power.
\[ (2^3)^2 = (2^3)(2^3) = (2 \cdot 2 \cdot 2)(2 \cdot 2 \cdot 2) = 2^6 \notag \]
The new power is 6, which just so happens to be the product of 2 and 3. Guess what? Ain't nothing special about the choice of base 2.
Let \(a, b,\) and \(c\) be real numbers. Then we have
\[ a^b a^c = a^{b+c} \quad \text{and} \quad \frac{a^b}{a^c} = a^{b-c} \quad \text{and} \quad (a^b)^c = a^{bc}. \notag \]
ALERT: The base \(a\) must be the same for this to work! Also, you must subtract the bottom power from the top power—order matters!
In caveman speak, "Multiplication in the base translates to addition in the exponent. Division in the base translates to subtraction in the exponent. Power to a power translates to multiplication in the exponent." You must memorize these rules!
Maybe you already noticed that \(2^1\) just turned out to be itself, based on the multiplication idea. That works for any base. Another special case is that any number raised to the power of zero is defined to be 1. Sound like black magic? Consider \( \frac{a^b}{a^b} = a^{b-b} = a^0 \) and look back at the beginning fraction. Whoop, it was the same on top and bottom. That's just 1!
- Let \(a\) be any real number. Then \( a^0 = 1\) and \( a^1 = a\).
- For any power \(n\), we have \( 0^n = 0 \) and \(1^n = 1 \).
Okay so we're feeling good about positive integers in exponents. What about negative integers? Well, we saw subtraction appear when we divided \( \dfrac{2^3}{2^2} = 2^{3-2} \). If we wrote \(2^{3-2} = 2^{3+(-2)} = 2^3 \cdot 2^{-2} \), it becomes clear that using a negative exponent means dividing by 2 a certain number of times! We can write division as a negative exponent, and we can write negative exponents as division, and as a consequence, we can move factors between numerators and denominators just by changing the sign of the exponent.
\[ 2^{-2} = \frac{1}{2^2}, \quad \quad \frac{3}{2^2} = 3 \cdot 2^{-2}, \quad \quad \frac{2^{3}}{3^{-2}} = 2^3 3^2 \notag \]
Let \(a\) and \(b\) be real numbers. Then \( a^{-b} = \dfrac{1}{a^b} \). Note that this also means \( \dfrac{1}{a^{-b}} = a^b \).
Let's shake things up and change some bases instead of exponents. Check it out:
\[3^2 \cdot 2^2 = 3 \cdot 3\cdot 2 \cdot 2 = 3 \cdot 2 \cdot 3 \cdot 2 = 6 \cdot 6 = 6^2. \notag \]
Same power on different bases being multiplied? Turns out you can multiply the bases and then raise it to the power! And look, it works backwards!
\[ (2\cdot 3)^2 = (2 \cdot 3)(2 \cdot 3)= 2^2 3^2 \notag \]
Let \(a, b, \) and \(c\) be real numbers. Then \( (ab)^c = a^c b^c \). ALERT: The power \(c\) must be the same to use this rule backwards.
You cannot pass an exponent into a set of parentheses with a plus or minus! Aka, \( (a+b)^c \neq a^c + b^c \). This is illegal!!!! Do not do it.
Now let's confront the elephant in the room: negatives. Consider \( (-2)^2 = (-2)(-2) = 4\). Squaring a negative number means multiplying a negative by a negative, which results in a positive number. Meanwhile, \( (-2)^3 = (-2)(-2)(-2) = (4)(-2) = -8 \) stays negative. Raising a negative number to an even power gives a positive result, and raising a negative number to an odd power gives a negative result.
Parentheses are extremely important here. Note that \( - 5^2 = - (5\cdot 5) = - 25 \) while \( (-5)^2 = (-5)(-5) = 25 \).
Let \(a\) be a positive base.
- When \( n\) is even, \( (-a)^n = a^n \).
- When \(n\) is odd, \( (-a)^n = - a^n \).
Let's bring it all together by doing some examples, and then you will try an exercise. We're going to start using some letters instead of numbers, so don't panic! Just push them around according to the rules and pretend they're your favorite numbers if it makes you feel better for now.
Simplify the expressions.
1. \( \dfrac{(-5)^3}{5^4} \)
2. \( 2^2 3^3 2^{-1} (2\cdot 3)^6 \)
3. \( ((-a)^2b^3)^4 \)
4.\( \dfrac{x^2 y^{-3}}{x^3 y} \)
Solution
1. \( \dfrac{(-5)^3}{5^4} = \dfrac{- 5^3}{5^4} = -\dfrac{5^3}{5^4} = -5^{3-4} = -5^{-1} =- \frac{1}{5^1} = - \frac{1}{5} \). You do not need to write all these steps out once these simplifications become second nature. I'm just trying to emphasize which rules are being used at each step.
2. I'm going to highlight in red the pieces I'm working on from step to step. Notice how these bases match: \( \textcolor{red}{2^2} 3^3 \textcolor{red}{2^{-1}} (2\cdot 3)^6 \). Bring them together to get \( \textcolor{red}{2^{2-1}} 3^3 (2 \cdot 3)^6 = \textcolor{red}{2^1} 3^3 (2 \cdot 3)^6\). Let's use the power of a product rule next: \( 2^1 3^3 \textcolor{red}{ (2 \cdot 3)^6 } = 2^1 3^3 \textcolor{red}{ 2^6 3^6 }\). Now we have a lot of matching bases we can combine by adding the exponents. Our final answer is \( 2^7 3^9 \). I won't ask you to calculate that by hand, it's a huge number. This is as clean as it gets!
3. Uh oh, letters! It's okay, we still follow the same rules. First note that \( (-a)^2 = a^2 \), so we actually have \( (a^2 b^3)^4 \). Then by the power of a product rule, we rewrite: \( (a^2)^4(b^3)^4 \). Then by the power to a power rule, we have: \(a^{2(4)}b^{3(4)}=a^8 b^{12} \).
4. We move everybody to the numerator by changing the sign of the exponent, and rewrite: \( \dfrac{x^2 y^{-3}}{x^3 y} = x^2 y^{-3} x^{-3} y^{-1} = x^{2-3}y^{-3-1} = x^{-1}y^{-4} = \dfrac{1}{xy^4} \).
Simplify.
- \( (2^3 \cdot (-2)^5) \)
- \( \frac{x^6}{(-x)^2} \)
- \( (3^4)^2 \)
- \( \frac{-2^4 \cdot 3^2}{2^2 \cdot 3^3} \)
- \( \frac{x^8 y^{-3}}{x^5 y^2} \)
- \( \left( \frac{a^5}{a^2} \right)^3 \left( \frac{b^4}{b^3} \right)^2 \)
- Answer
-
- \( - 2^8 \)
- \( x^4 \)
- \( 3^8 \)
- \( - \frac{2^2}{3} = \frac{4}{3} \)
- \( \frac{x^3}{y^5} \) or \(x^3 y^{-5} \) is fine
- \(a^9 b^2\)
Working With Radicals (Roots)
Here's a question... If there was such thing as a fractional power, say like \( 4^{\frac{1}{2}} \), the power rules would imply that
\[ ( 4^{\frac{1}{2}})^2 = 4^{\frac{1}{2} \cdot 2} = 4^1 = 4. \notag \]
In other words, \( 4^{\frac{1}{2}} \) should be a number that, when squared, results in \(4\). Who do we know who works like that? (Dora the Explorer pause here.)
Did you say \(2\)? Nice. Did you say \(-2\) too? You iconoclast! We say that \(2\) and \(-2\) are square roots of \(4\). You will have seen this written \(\sqrt{4} = \pm 2 \), and now we know that the root symbol can be written using a fractional power, \(\sqrt{4} = 4^{\frac{1}{2}} \). When you're given the radical symbol \( \sqrt{ } \) in a problem, by convention it is defined to report the positive square root (the principal root).
If \( a^2 = b \), then \(a\) is a square root of \(b\). For \( a \geq 0 \), we write \( \sqrt{b} = a \).
You can ask the same question about \( 8^{\frac{1}{3}} \), whose answer would be \(2\) since \(2^3 = 8\), and we write \( \sqrt[3]{8} = 2 \), saying "2 is the cube root of 8."
If \(a^n = b \), then \(a\) is an nth root of \(b\). The principal nth root of \(b\) is written \( \sqrt[n]{b} = a \), where \(a \geq 0\).
Convince yourself of the following fun fact.
For any \(n\), we have \( \sqrt[n]{0} = 0 \) and \( \sqrt[n]{1} = 1 \).
You will want to get really good at writing roots as fractional powers and combining them with exponents, because you do it all the time in Calculus I! So here's that stuff.
We write \(n\)th roots as fractional powers, \(\sqrt[n]{a} = a^{\frac{1}{n}} \), and we have
\[ (\sqrt[n]{a})^m = \left(a^{\frac{1}{n}} \right)^m = a^{\frac{m}{n}} = \sqrt[n]{a^m}. \notag \]
Notice that line is saying that it doesn't matter if you take a root first and then a power, or take the power first and then the root! It all follows from the power to a power rule. All you have to do is place any powers into the numerator of a fractional exponent, and place any roots into its denominator.
Simplify.
1. \( (\sqrt[3]{a^2})^6 \)
2. \( \sqrt{ (a^3 \sqrt{b})} \)
Solution
1. Looking at the inside first, we have \( (\sqrt[3]{a^2})^6 = (a^{\frac{2}{3}})^6\). Then using power rules, we have \( (a^{\frac{2}{3}})^6 = a^{\frac{2\cdot 6}{3}} = a^4 \).
2. We convert the outer square root into a fractional power first, so \( \sqrt{ a^3 \sqrt{b}} = (a^3 \sqrt{b})^{\frac{1}{2}} \). We bring it through the product to get \( (a^3)^{\frac{1}{2}} (\sqrt{b})^{\frac{1}{2}} = a^{\frac{3}{2}} (b^{\frac{1}{2}})^{\frac{1}{2}} = a^{\frac{3}{2}} b^{\frac{1}{4}} \).
Regarding negatives under radicals, we run into some legality issues. For example, any number will give a positive result when it gets squared, right? So there's no way to have a number who gets squared to produce a negative result. This means there is no real number answer to \(\sqrt{-4}\). (There is a world beyond the real numbers, but we don't live there right now.) On the other hand, it does make sense to ask about \( \sqrt[3]{-8}\), because we know \( (-2)^3 = -8 \), so that'll work as the answer!
You are not allowed to take even roots ( \(\sqrt{}, \sqrt[4]{}, ... \) ) of negative numbers! But you can take odd roots ( \( \sqrt[3]{}, \sqrt[5]{}, ... \) ) of negative numbers, and the result will be negative.
First practice computing some roots:
Simplify (without using a calculator).
- \( \sqrt{16} \)
- \( \sqrt[3]{125} \)
- \( \sqrt{3^2} \)
- \( \sqrt[3]{-27} \)
- \( \sqrt[8]{1} \)
- \( \sqrt{400} \)
- Answer
-
- This is 4.
- This is 5.
- \( \sqrt{3^2} = \sqrt{9} = 3 = (\sqrt{3})^2 \).
- This is 1.
- Notice that \(400 = 20*20 = 20^2 \) so this is 20.
In working with radicals, we need a very important rule.
You can break up a root over a multiplication or division:
\[ \sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b} \quad \text{and} \quad \sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \notag \]
These rules also work backwards!
You cannot break up a root over a plus or minus! Aka, \( \sqrt{a + b} \neq \sqrt{a} + \sqrt{b} \). It is a lie. Straight to jail.
Whenever you take a square root of some number \(a\) and the answer is a whole number, we call \(a\) a perfect square. Examples would be 4, 9, 16, 25, etc. Sometimes, a root doesn't work out nice and clean, but you can still do a little simplifying. This utilizes the rule above. For example, \( \sqrt{32} \) isn't a whole number, but we can write
\[ \sqrt{32} = \sqrt{16\cdot 2} = \sqrt{16}\sqrt{2} = 4 \sqrt{2}. \notag \]
It all boils down to finding any perfect squares under the square root, like the 16 hidden inside the 32. Another aspect to cleaning up expressions with radicals in them is the topic of "rationalizing the denominator." I don't really believe in it personally (we haven't needed it since we stopped approximating radicals by hand like peasants), but it's the process of multiplying top and bottom of a fraction by whatever is needed to remove radicals from the denominator. For example,
\[ \frac{3}{\sqrt{5}} = \frac{3 \cdot \sqrt{5} } {\sqrt{5} \cdot \sqrt{5}} = \frac{3\sqrt{5}}{5} . \notag \]
Simplify.
1. \( \sqrt{\frac{1}{25}} \)
2. \( \sqrt[3]{x^6} \)
3. \( \frac{ \sqrt[4]{16}}{ \sqrt{4}} \)
4. \( \frac{ \sqrt{2} + \sqrt{18}}{2} \)
5. \( \sqrt{ x^3 y^4 z^5} \)
6. \( (\sqrt[4]{ab^2})^8 \)
Solution
1. \( \sqrt{\frac{1}{25}} = \frac{\sqrt{1}}{\sqrt{25}} = \frac{1}{5} \). You don't have to write the middle step if you don't want to.
2. Using power rules, \( \sqrt[3]{x^6} = x^{\frac{6}{3}} = x^2 \).
3. \( \frac{ \sqrt[4]{16}}{ \sqrt{4}} = \frac{2}{2} = 1 \)
4. Cleaning up that \(\sqrt{18}\) as much as possible, \( \frac{ \sqrt{2} + \sqrt{18}}{2} = \frac{ \sqrt{2} + \sqrt{2 \cdot 3^2}}{2} = \frac{ \sqrt{2} + 3\sqrt{2} }{2} \). Then we combine the single \( \sqrt{2} \) with the other three of 'em in the numerator to get \( \frac{ 4\sqrt{2} }{2} = 2 \sqrt{2} \).
5. Collecting as many perfect squares as we can and bringing them outside the radical sign, \( \sqrt{ x^3 y^4 z^5} =\sqrt{ x \cdot x^2 \cdot (y^2)^2 \cdot z \cdot (z^2)^2} = xy^2z^2 \sqrt{xz} \)
6. Using fractional powers, rewrite: \( (\sqrt[4]{ab^2})^8 = (ab^2)^{\frac{8}{4}} = (ab^2)^{2} \). Then using power rules, simplify to \( a^2 b^4 \).
Now you try the exercises in the next section!