Lesson Library> Exponents & Radicals
Exponents & Radicals
Introduction to Exponents
The square below has an area of 5. What is the length of the square’s side x?
A square with side length 2 has an area of \(2 \cdot 2 = 4\) and a square with a side length 3 has an area of \(3 \cdot 3 = 9\). The square above must therefore have a side length that is somewhere between 2 and 3. But what is that number exactly?
How Exponents Work
Thanks to multiplication, we can shorten sums that only contain the same term. For example, \(4 + 4 + 4 + 4 + 4 = 4 \cdot 5\) and \(x + x + x = 3x\).
There is also a system for shortening products that only contain the same factor. If we would like to multiply 7 by itself four times, we can write this as \(7 \cdot 7 \cdot 7 \cdot 7 = 7^4\). Similarly, \[ x \cdot x \cdot x \cdot x \cdot x \cdot x = x^6.\]
An expression like \( 7^4 \) is called a power. The 7 is called the base and the 4 is called the exponent of the power. The entire expression is read as "seven to the power of four", or just “seven to the four”.
When a number is raised to the power of 2 we often say that the number is squared. This is because the area of a square with side length x has the area \( x \cdot x = x^2 \). Similarly, when a number is raised to the power of 3 we say that the number is cubed
Exercise 1
Write the following products and phrases as powers.
- “Nine cubed”
- \( 1000 \cdot1000 \cdot1000 \cdot1000 \cdot1000 \)
- \( (-1) \cdot (-1) \cdot (-1) \cdot (-1) \cdot (-1) \cdot (-1) \cdot (-1) \)
- "Fourteen to the power of five"
- \( \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \)
- \( 5 \)
Answers:
- \( 9^3 \)
- \( 1000^5 \)
- \((-1)^7 \)
- \( 14^5 \)
- \( \left( \frac{1}{2} \right)^4 \)
- \( 5^1 \) or just \( 5 \)
Exponents and Order of Operations
In our previous lesson on order of operations, we explained why operations such as multiplication, division, addition and subtraction must be performed in a certain order to get the correct result.
In \(2 + 5 \cdot 3\), for example, we must first multiply 5 and 3 before adding 2. If we add 2 and 5 first and then multiply by 3 we will multiply 3 by 7 and not by 5 as the expression indicates. To avoid one operation affecting another, we always perform all multiplications and divisions before moving on to addition and subtraction.
The same applies when using exponents. Let’s take \(4 \cdot 5^3\) as an example. This expression says that we should take 5 raised to 3 and multiply the result by 4. We can also write this as \(4 \cdot 5 \cdot 5 \cdot 5\). What happens if we instead multiply 4 by 5 first and then take the result of this to the power of 3? We would get \( 20^3 = ( 4 \cdot 5 )^{3} = 4^{3} \cdot 5^{3}\). Which is obviously not the same as \( 4 \cdot 5^3 \).
This happened because the multiplication "spilled over" and affected the exponent operation. This is the same problem as with addition and multiplication. To correctly determine the value of this expression, we must first perform all exponent operations before any multiplication or division can be done.
If we want to perform any operation before we start with the exponents, we must put that operation in brackets. In the example above we would then write \( 4 \cdot 5)^3 \). Brackets always have priority over all other operations. We can summarise the order in which operations must be performed with the following list:
Order of Operations
- Brackets
- Exponents
- Division and Multiplication
- Addition and Subtraction
A common trick for remembering this order is the abbreviation BIDMAS, which stands for:
- Brackets
- Indices
- Division
- Multiplication
- Addition
- Subtraction
“Indices” is another word for “exponents”.
Exercise 1
- \(3^2 \cdot 7 \)
- \(6 + 2^3 \cdot 9 \)
- \(2^3 \cdot 2^4 + 2 \cdot3 - 2^5 \)
- \( (6 + 2 \cdot 2)^2 \)
- \( \cfrac{5^3 + 5}{25} \)
- \( \cfrac{3^3 \cdot 3^5}{( 1 + 2^3)^2} \)
a) $3^2 \cdot 7 = 9 \cdot 7 = \mathbf{63} $
b) $6 + 2^3 \cdot 9 = 6 + 8 \cdot 9 = 6 + 72 = $$ \ \mathbf{78} $
c) $2^3 \cdot 2^4 + 2 \cdot 3 - 2^5 = $$ \ 8 \cdot 16 + 2 \cdot 3 - 32 = 128 + 6 - 32 = $$ \ \mathbf{102} $
d) $(6 - 2 \cdot 2)^4 = (6 - 4)^4 = $$ \ 2^4 = $$ \ \mathbf{16} $
e) $\dfrac{5^3 + 5}{25} = \dfrac{125 + 5}{25} = $$ \ \dfrac{130}{25} = $$ \ \mathbf{\dfrac{26}{5}} $
f) $\dfrac{3^3 \cdot 3^5}{(1 + 2^3)^2} = \dfrac{3^8}{(9)^2} = $$ \ \dfrac{3^8}{( 3^2)^2} = $$ \ \dfrac{3^8}{3^4} = 3^4 = $$ \ \mathbf{81} $
Exercise 2
What is the difference between simplifying $\dfrac{6^4 \cdot 6^2}{6}$ and simplifying $\dfrac{6^4 + 6^2}{6}$?
In the first expression we are only multiplying factors. We can actually view dividing by $6 $ as if we are multiplying by $\frac{1}{6} $:
$$ 6^4 \cdot 6^2 \cdot \dfrac{1}{6} $$
In a product, it doesn’t matter in which order the factors are multiplied. For example, to make the entire product above three times larger, we can choose between either multiplying the first factor by $3 $, the second factor by $3 $ or the third factor by $3$. Whichever of the factors we choose does not matter.
The same does not apply to a sum of terms. In order for the sum to be three times larger, all terms must be multiplied by $3 $. Similarly, all terms in a sum must be divided by $3 $ if we want to reduce the sum to a third of its size.
This is the key difference between the two expressions above. When simplifying $\dfrac{6^4 \cdot 6^2}{6}$ we only need to divide one of the factors by $6$. But when simplifying $\dfrac{6^4 + 6^2}{6}$ we need to divide all terms by $6$.
Negative Bases
What happens when the base is a negative number, such as $(-2)^4 $? We just do the same as before and multiply $(-2) \cdot (-2) \cdot $$(-2) \cdot (-2) =$$ 16 $. A negative number raised to an even exponent always gives a positive result since the product of two negative numbers will be positive. If the exponent is odd, however, the result will be negative. For example, $(-1)^{1000} = 1 $ while $(-1)^{1001} = -1 $.
Exercise 3
Determine the value of the following powers.
a) $- (2)^3 $
b) $(- 1)^{100} $
c) $- (- 3)^2 $
a) $- (2)^3 = - (2 \cdot2 \cdot2) = \mathbf{-8} $
b) $(- 1)^{100} = 1^{100} = \mathbf{1} $
c) $-(-3)^2 = $$ \ -((-3) \cdot (-3)) = - (9) = \mathbf{-9} $
Exercise 4
Is $(-2)^4 $ the same as $-(2^4)$?
The first expression means $(-2) \cdot (-2) \cdot (-2) \cdot (-2) =$$ \ 16 $. The second means $- (2 \cdot2 \cdot2 \cdot2) \ = $$ \ -16 $. If the minus sign is inside the brackets, i.e the base of the power is negative, we should multiply the negative base times by itself as many times as the exponent indicates. But if the minus sign is outside the brackets the base is positive. We then first determine the value of the power and then we simply change the sign of the result.
When the minus sign is inside the brackets, the result can be either positive or negative depending on whether the exponent is odd or even. If the minus sign is outside the brackets, the result is always negative when the base is a positive number.
Exponent Rules
Adding and Subtracting Exponents
Let's say we want to multiply $2^3 \cdot2^5 $. The first power is the product of three $2 $s and the second power is the product of five $2 $s. The final product then contains eight $2 $s, so $2^3 \cdot2^5 =$$ 2^8 =$$ \mathbf{256}$.
Exercise 1
Write the following products as a single power.
a) $9^4 \cdot 9^8 = 9^{4 + 8} = \mathbf{9^{12}} $
b) $x^5 \cdot x^{13} = x^{5 + 13} = \mathbf{x^{18 }} $
c) $4^{16} \cdot 4^6 \cdot 4^{21} = 4^{16 + 6 + 21} = \mathbf{4^{43}} $
d) $\dfrac{8^{20}}{8^{11}} = 8^{20-11} = \mathbf{8^{9}} $
e) $\dfrac{13^{3} \cdot13^{28}}{13^{7}} = 13^{3 + 28-7} = \mathbf{13^{24}} $
f) $\dfrac{y^{19} \cdot y^{ 34}}{y^{5} \cdot y^{12}} = y^{19 + 34-5-12} = \mathbf{y^{36}} $
Exercise 2
How many times larger is $3^{87} $ compared to $3^{85} $?
$3^{87} $ is the same as $3^2 \cdot 3^{85} = 9 \cdot 3^{85} $. This means that $3^{87} $ is $\mathbf{9}$ times larger than $3^{85}$.
Exercise 3
Which expression is twice as large as $2^{4x + 5} $?
To get an expression that is twice as large, we only need to multiply by $2 $. The answer is thus $2 \cdot 2^{4x + 5} = 2^{4x + 5 + 1} = \mathbf{2^{4x + 6}} $.
Multiplying Bases
Let us now examine the expression $3^5 \cdot 4^5 $. Can we simplify this in any way? If we write out the entire product we get:
$ (3 \cdot3 \cdot3 \cdot3 \cdot3) \ \cdot \ (4 \cdot4 \cdot4 \cdot4 \cdot4) $
If we rearrange the numbers, we can create five pairs of threes and fours:
$ (3 \cdot4) \ \cdot (3 \cdot4) \ \cdot (3 \cdot4) \ \cdot (3 \cdot4) \ \cdot (3 \cdot4) $
Which is the same as $12 \cdot12 \cdot12 $$ \ \cdot12 \cdot12 $ or $12^5 $. If the powers have the same exponent, we can just multiply the bases with each other. The final product still consists of the same factors, and since it does not matter in which order they are multiplied, we end up with the same result. This gives us another useful rule for managing powers:
$$\mathbf{a^n \cdot \ b^n = (ab)^{n}}$$
Exercise 4
Write the following products as a single power with the lowest possible base.
a) $2^{10} \cdot7^{10} $
b) $4^4 \cdot 3^4 \cdot 6^4 $
c) $a^{2x} \cdot b^{2x} \cdot c^{2x}$
a) $2^{10} \cdot7^{10} = (2 \cdot 7)^{10} = \mathbf{14^{10}} $
b) $4^4 \cdot3^4 \cdot6^4 = (4 \cdot 3 \cdot 6)^4 = \mathbf{72^4} $
c) $a^{2x} \cdot b^{2x} \cdot c^{2x} = $$ \ \mathbf{(abc)^{2x}}$
Exercise 5
How many times larger is $\dfrac{12^{101} \cdot 7^{100}}{3^{99}} $ compared to $\dfrac{12^{100} \cdot 7^{99}}{3^{98}} $?
We can rewrite $\dfrac{12^{101} \cdot 7^{100}}{3^{99}} $ to:
$$ \dfrac{12 \cdot 7}{3} \cdot \dfrac{12^{100} \cdot 7^{99}}{3^{98}} \ = \ 28 \cdot \dfrac{12^{100} \cdot 7^{99}}{3^{98}} $$
So the first expression is $\mathbf{28} $ times larger than the second expression.
Multiplying Exponents
What is $(2^3)^4 $? We are first taking $2 $ to the power of $3 $. Then we are taking the result of this to the power of $4 $. The first step gives us three $2 $s. The second step then gives us four groups with three $2$s in each. This gives us a total of $3 \cdot 4 = \mathbf{12}$ $2 $s.
$ (2^3)^4 = $$ \ (2 \cdot 2 \cdot 2) \ \cdot $$ \ (2 \cdot 2 \cdot 2) $$ \ (2 \cdot 2 \cdot 2) \ \cdot \ (2 \cdot 2 \cdot 2) =$$ \ 2^{3 \ \cdot \ 4} = $$ \ 2^{12} $
This example gives us another useful rule. When a number raised by a certain exponent is raised to an additional exponent, we multiply the two exponents by each other. It is generally true that
$$ \mathbf{(a^m)^n = a^{m \ \cdot \ n}} $$
Exercise 6
Write the following expression to a single power.
a) $(5^{10})^7$
b) $((8^{4})^2)^3$
c) $(((x^d)^a)^c)^b$
a) $(5^{10})^7 = $$ \ 5^{10 \ \cdot \ 7} = $$ \ \mathbf{5^{70}}$
b) $((8^{4})^2)^3 = $$ \ 8^{4 \ \cdot \ 2 \ \cdot \ 3} = $$ \ \mathbf{8^{24}}$
c) $(((x^d)^a)^c)^b = $$ \ \mathbf{x^{abcd}}$
Exercise 7
Is $(2^3)^4 $ the same as $2^{3^4} $?
The first expression is telling us to first take $2 $ to the power of $3 $ (operations in brackets always come first) and then to take the result of this to the power $4 $.
The second expression, however, is telling us to first take $3 $ to the power of $4 $ and then take $2 $ to the power of the result of $3^4 $.
When determining the value of these expressions we get $(2^3)^4 = 2^{3 \cdot 4} = 2^{12} $ while $2^{3^4} = 2^{3 \cdot 3 \cdot 3 \cdot 3} = 2^{81} $. So these expressions are not the same.
Exercise 8
What is $\dfrac{2^{2^{2^2}}}{((2^2)^2)^2}$?
In the numerator, the expression is telling us to take $2$ to the power of $2^{2^2}$. Before we can perform this operation we must first figure out what the exponent $2^{2^2}$ is. And before we can figure out what $2^{2^2}$ is, we must first solve that this exponent's exponent is equal to $2^2=4$. We always start with the innermost exponent and then work our way outwards from there.
$\dfrac{2^{2^{2^2}}}{((2^2)^2)^2} \ =$
$\dfrac{2^{2^4}}{2^{2 \ \cdot \ 2 \ \cdot \ 2}} \ =$
$\dfrac{2^{16}}{2^8} \ =$
$2^{16-8} \ =$
$2^{8} \ =$
$\mathbf{256}$
Negative Exponents
If we multiply a power by its base, the exponent increases by $1 $. For example, $5^3 \cdot5 = 5^4 $. In the same way, the exponent decreases by $1 $ if we divide the power by its base.
What happens if we continue to divide a power with its base over and over again? The exponent will then continue to decrease by $1 $ until it becomes $0$ and then turns negative.
$ \ \ \ 5^3 \ \ \ \ \ \ \ 5^2 \ \ \ \ \ \ \ 5^1 \ \ \ \ \ \ 5^0 \ \ \ \ \ \ 5^{- 1} \ \ \ \ \ \ 5^{- 2} \ \ \ \ \ \ 5^{- 3} $
$ 125 \ \ \ \ \ \ 25 \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \dfrac{1}{5} \ \ \ \ \ \ \ \dfrac{1}{25} \ \ \ \ \ \ \dfrac{1}{125} $
Just as $5^3 $ means that we have multiplied $5 $ three times, $5^{- 3} $ means that we have divided by $5 $ three times. It is generally true that:
$$ \mathbf{a^{- n} = \dfrac{1}{a^n}}$$
When determining the value of power with negative exponents, it's sometimes easiest to invert the power first so the exponent becomes positive. Then we can determine the value of the power as we normally do.
Exercise 1
Determine the following powers.
a) $5^{- 2} $
b) $3^{- 4} $
c) $(- 2)^{- 3} $
d) $\Big( \dfrac{1}{ 2} \Big)^{- 1} $
e) $- \Big( \dfrac{2}{3} \Big)^{- 2} $
a) $5^{- 2} = $$ \dfrac{1}{5^2} = $$ \ \mathbf{\dfrac{1}{25}}$
b) $3^{- 4} = $$ \dfrac{1}{3^4} = $$ \ \mathbf{\dfrac{1}{81}}$
c) $(- 2)^{- 3} = $$ \dfrac{1}{(-2)^3} = $$ \ \mathbf{- \dfrac{1}{8}}$
d) $\Big( \dfrac{1}{ 2} \Big)^{- 1} = $$ \ (2^{-1})^{-1} = $$ = \ 2^{(-1 \ \cdot \ -1)} \ = $$ \ \mathbf{2}$
e) $- \Big( \dfrac{2}{3} \Big)^{- 2} = $$ \ - \dfrac{1}{\Big( \dfrac{2}{3} \Big)^2} = $$ \ - \Big( \dfrac{3}{2} \Big)^2 = $$ \ \mathbf{- \dfrac{9}{4}}$
Exercise 2
Write the following numbers as powers form with the smallest possible integer as base.
a) $64 $
b) $- \dfrac{1}{5} $
c) $100 $
e) $\dfrac{1}{4} $
f) $- \dfrac{1}{27} $
a) $\mathbf{2^6}$
b) $\mathbf{5^{- 1}}$
c) $\mathbf{10 ^ 2}$
d) $\mathbf{2^{- 2}}$
e) $\mathbf{- 3^{- 3}}$
As you can see, $a^0 = 1 $ because $\dfrac{a^1}{a} = a^0 = 1 $. Any number raised to $0 $ always becomes $1 $.
There is only one exception to this rule and that is if $a = 0 $. In this case, the reasoning $\dfrac{0^1}{0} = 0^0 $ because we may never divide a number by $0 $. The expression $0^0 $ is therefore called undefined as we can’t assign a specific value to it. It is generally true that:
$$\mathbf{a^0 = 1} \ \ \ a ≠ 0$$
Exercise 3
What is the value of
$ 4^6 \cdot 2^{- 24} \cdot 2^5 \cdot $$ \ \dfrac{1}{16} \cdot 8^5 \ ? $
All the numbers can be written as powers with base $2 $. This gives us:
$ 4^6 \ \cdot \ 2^{- 24} \ \cdot $$ \ 2^5 \ \cdot \ \dfrac{1}{16}$$ \ \cdot \ 8^5 \ =$
$ (2^2)^6 \ \cdot \ 2^{- 24} \ \cdot $$ \ 2^5 \ \cdot \ 2^{-4}$$ \ \ \cdot (2^3)^5 \ =$
$ 2^{12 -24 + 5 -4 + 15} \ \ =$
$ 2^{4} =$
$ \mathbf{16} $
Scientific Form
In science, it’s common for us to work with numbers that can be both extremely large or microscopically small. In some cases, they may be so large or so small that our normal way of writing numbers is not practical for describing them. Try to get an idea of how far it is to the nearest star ($40$ $208$ $000 $ $000$ $000$ kilometers) or how small the mass of a water molecule is ($0.000$ $000$$000$ $000$ $000$ $000$ $000$ $029$ $88$ grams). To solve this problem, such extreme numbers are usually rewritten in so-called scientific form.
Scientific form means that a number is written on the form:
$$k \cdot 10^n $$
where $1<k<10$ and $n$ is an integer.
Exercise 1
How can we determine how large the base $10$ exponent should be when writing a number in scientific form?
Let's take the number $40$ $208$ $000$ $000$ $000 $ as an example. To determine how many $10 $s are needed, we can imagine putting our pencil tip just behind the $4 $ and then counting the number of $10$ jumps we need to make backwards to get to the starting point. In this case, we first need to make one jump back and then jump four blocks of three zeros. In total, we need to make $1 + 4 \cdot3 = 13 $ such jumps, so the exponent of the number in scientific form is $\mathbf{13} $.
We can use the same method for the number $0.000$ $000$$000$ $000$ $000$ $000$ $000$ $029$ $88$. Putting our pencil tip just behind the first value digit $(2)$ and counting how many forward-jumps we have to make to get just above the decimal point. In this case, we first must make two forward-jumps and then jump seven blocks of three zeros. In total, we need to make $2 + 3 \cdot7 = 23 $ such jumps, so the exponent of the number in scientific form is $\mathbf{-23} $.
Exercise 2
Write the following numbers in scientific form.
a) $50 $
b) $0.007$
c) $351$ $000$ $000 $
d) $0.000$ $000$ $000$ $006$ $192$
e) $447.2 \cdot 10^{29} $
f) $10273 \cdot 10^{- 15} $
g) $0.00146 \cdot 10^{54} $
h) $\dfrac{12}{240 \ 000 \ 000 \ 000 \ 000} $
a) $\mathbf{5 \cdot10^1} $
b) $\mathbf{7 \cdot10^{- 3}} $
c) $\mathbf{3.51 \cdot 10^8} $
d) $\mathbf{6.1922 \cdot 10^{- 12}} $
e) $447.2 \cdot 10^{29} \ =$$ \ (4.472 \cdot 10^{2}) \cdot 10^{29} \ =$$ \ \mathbf{4.472 \cdot 10^{ 31}} $
f) $10273 \cdot 10^{- 15} \ =$$ \ (1.0273 \cdot 10^4) \cdot 10^{- 15} \ =$$ \ \mathbf{1.0273 \cdot 10^{- 11}} $
g) $0.00146 \cdot 10^{54} \ =$$ \ (1.46 \cdot 10^{- 3}) \cdot10^{54} \ =$$ \ \mathbf{1.46 \cdot 10^{51}} $
h) $\dfrac{12}{240 \ 000 \ 000 \ 000\ 000} \ \ \ =$$ \ \dfrac{12}{24} \cdot \dfrac{1}{10 \ 000 \ 000 \ 000 \ 000} \ \ \ =$$ \ 0.5 \cdot 10^{- 13} \ =$$ \ \mathbf{5 \cdot10^{- 14}} $
Exercise 3
A human brain cell weighs on average $1 \cdot 10^{- 8} $ grams while the mass of the planet is approximately $6 \cdot10^{21} $ metric tons ($1$ ton = $1000$kg). How many times larger is the mass of the earth compared to the mass of an average human brain cell?
Note that the mass of the earth is expressed in tons and not in grams. To compare the size of any two objects by scientific form, we must first express them with the same unit. One metric ton is equal to $1000 $ kg and $1$ kg is equal to $1000$ grams. A ton is thus equal to $1000 \cdot1000 = 10^6 $ grams, so the mass of the earth is $6 \cdot10^{21} \cdot 10^6 = 10^{27} $ grams.
To figure out how many times greater $6 \cdot10^{27} $ is compared to $1 \cdot 10^{- 8}$, we need to divide the first number by the second.
$$ \dfrac{6 \cdot10^{27}}{1 \cdot 10^{- 8}} \ = $$
$$\cdot 6 \cdot 10^{27-(-8)} \ =$$
$$6 \cdot 10^{35} $$
The mass of the earth is therefore $\mathbf{6 \cdot 10^{35}} $ times larger than with the mass of an average human brain cell.
Fractional Exponents and Radicals
We have seen how exponents can be either positive or negative. But the journey does not end there! We can also use exponents that are not whole numbers.
Let’s take the power $3^{1/2} $ as an example. What is the value of this number? We know that $a^m \cdot \ a^n =$$ \ a^{m \ + \ n} $. It must therefore be true that:
$3^{1/2} \cdot \ 3^{1/2} \ = \ 3^1 \ = \ 3 $
$3^{1/2} $ is the number that multiplied by itself equals $3 $.
By trial and error, we can conclude that $3^{1/2} ≈ 1,732 $. The decimals in $3^{1/2} $ never end, so we can’t write the complete number even if we wanted to. Therefore, we just write $3^{1/2} $ and leave it at that.
It’s very common to work with powers that have fractional exponents like $3^{1/2}$. This expression has therefore been given its own symbol:
$$3^{1/2} \ = \ \mathbf{\sqrt{3}}$$
The $\sqrt{ \phantom{x}}$ symbol is called a radical and $\sqrt{3} $ as is read as "the square root of $3 $". Thus, $\sqrt{3} \cdot \sqrt{3} = 3 $.
We can also choose to write $5^{1/3} $ as $\sqrt[3]{5} $, which reads as "the cube root of $5$". We put a $3$ at the root sign to indicate that we must multiply $\sqrt[3]{5} $ by itself three times to get $5 $.
Similarly, $10^{1/4} = \sqrt[4]{10} $ and $16^{1/7} = \sqrt[7]{ 16} $. These are read as "the fourth root of $10 $" and "the seventh root of $16$".
It is generally true that:
$$\mathbf{a^{1 / n} = \sqrt[n]{a}}$$
and
$$\mathbf{\sqrt[n]{a^n} = a }$$
The number underneath the root sign is called the radicand. When we add two or more roots with the same radicand, this is written as $\sqrt{3} + \sqrt{3} = 2 \sqrt{3}$.
Exercise 1
Let's return to the question we posed at the beginning of the lesson:
The square below has an area of $5$. What is the length of the square’s side $x$?
The area of a square is the product of its sides. This means that:
$$x \cdot x \ = \ 5$$
$$x^2 \ = \ 5$$
We can eliminate the exponent by taking the square root of both sides:
$$\sqrt{x^2} \ = \ \sqrt{5}$$
$$\mathbf{x \ = \ \sqrt{5}}$$
So the side of the square is $\mathbf{\sqrt{5} \approx 2.236}$.
Exercise 2
Determine the value of the following expressions.
a) $\sqrt{25} $
b) $\sqrt[3]{64} $
c) $10000^{1/4} $
d) $0^{1/2} $
e) $8^{1/3} $
a) $\mathbf{5}$
b) $\mathbf{4}$
c) $\mathbf{10}$
d) $\mathbf{0}$
e) $\mathbf{2}$
You may have already realized that $3 $ is not the only number that times itself becomes $9 $. The number $-3 $ satisfies this condition as well. Which answer should we then give when asked to determine $\sqrt{9}$? In this case it’s always the positive root that is implied.
A good rule of thumb is to give the positive solution if the root sign was included in the problem from the start. If the problem only included exponents, we state both the positive and the negative solution if one exists.
Example 1
What is the difference between $x^2 = 25 $ and $x = \sqrt{25} $?
Solution
$x^2 = 25 $ implies that we should find all numbers that multiplied by themselves equals $25 $, so $x=5$ or $x=-5$. We usually write this as $x = \pm 5 $.
$x = \sqrt{25} $ implies that we are only looking for non-negative solutions, so $x = 5 $. If the root sign was already included in the problem from the start, only the the positive solution is asked for.
So far we have only looked at powers where the numerator in the fractional exponent is $1 $. But we can also have powers like $8^{2/3} $. This is a bit more complex, but we can solve it by using the exponent rules we have covered above.
$ 8^{2/3} = 8^{1/3 \ \cdot \ 2} \ = $$ \ (8^{1/3})^2 = (2)^2 = 4 $
We start by separating the numerator and the denominator. The fractional exponent $2/3$ is thereby rewritten as a product $(1/3 \ \cdot \ 2)$.
Next, we take the base to the power of the exponent’s denominator $(1/3)$. Finally, we take the result of this to the power of the exponent’s numerator $(2)$. With a little practice, this method will work like clockwork for you.
Exercise 3
Determine the value of the following powers.
a) $25^{3/2} $
b) $8^{5/3} $
c) $\sqrt[4]{16^3} $
d) $100^{- 3/2} $
e) $\Big(\dfrac{1}{81} \Big)^{- 3/4} $
a) $25^{3/2} = 25^{1/2 \ \cdot \ 3} \ =$$ \ (25^{1/2})^3 = (5)^3 =$$ \ \mathbf{125} $
b) $8^{5/3} = 8^{1/3 \ \cdot \ 5} \ =$$ \ (8^{1/3})^5 = (2)^5 =$$ \ \mathbf{32} $
c) $\sqrt[4]{16^3} = (16^3)^{1/4 } \ =$$ \ 16^{3 \ \cdot \ 1/4} \ = (16^{1/4})^{3} \ = 2^3 =$$ \ \mathbf{8} $
d) $100^{- 3 / 2} = 100^{1/2 \ \cdot (-3)} \ =$$ \ (100^{1/2})^{- 3} \ = 10^{- 3} = \dfrac{1}{10^3 } =$$ \ \mathbf{\dfrac{1}{1000}} $
e) $\Big(\dfrac{1}{81} \Big)^{- 3/4} \ = (81^{- 1})^{- 3/4} \ =$$ \ ( 81)^{3/4} \ = (81)^{1/4 \ \cdot \ 3} \ = \ $$ \ (81^{1/4})^3 = (3)^3 =$$ \ \mathbf{27} $
Exercise 4
Determine the value of the following expressions.
a) $\Big(\dfrac{1}{8} \Big)^{2/3} $
b) $(\sqrt{1000})^{2/3} $
c) $( \sqrt[3]{2})^6 $
a) $\Big(\dfrac{1}{8} \Big)^{2/3} =$$ \ \dfrac{1^{2/3} }{8^{2/3}} =$$ \ \dfrac{1}{8^{1/3 \ \cdot \ 2}} =$$ \ \dfrac{1}{(8^{1/3})^2} =$$ \ \dfrac{1}{2^2} =$$ \ \mathbf{\dfrac{1}{4}} $
b) $(\sqrt{1000})^{2/3} =$$ \ (1000^{1/2 })^{2/3} =$$ \ 1000^{1/2 \ \cdot \ 2/3} =$$ \ 1000^{1/3} =$$ \ \mathbf{10} $
c) $3 \sqrt{2^6 } =$$ \ 3 (2^{6})^{1/3} =$$ \ 3 (2^{6 \ \cdot \ 1/3}) =$$ \ 3 (2^2) =$$ \ 3 \cdot 4 =$$ \ \mathbf{12} $
Exercise 5
Determine the value of the following expressions:
a) $9 \cdot \dfrac{1}{27} \cdot \sqrt{3}$
b) $7^{1/3} \cdot \sqrt[3]{7^2} \cdot 7^{- 2}$
a) $3^2 \cdot 3^{- 3} \cdot 3^{1/2} \ =$$ \ 3^{- 1/2} \ =$$ \ \mathbf{\dfrac{1}{\sqrt{3}}} $
b) $7^{1/3} \cdot \sqrt[3]{7^2} \cdot 7^{- 2} \ =$$ \ 7^{1/3} \cdot (7^2)^{1/3} \cdot 7^{- 2} \ =$$ \ 7^{1/3} \cdot 7^{2/3} \cdot 7^{ -2} \ =$$ \ 7^{- 1} \ = \ \mathbf{\dfrac{1}{7}} $
Exponent Rules - Summary
We have now covered many useful rules for managing exponents. Here is a complete summary of all the important rules you should know:
Adding and subtracting exponents
$a^m \cdot \ a^n = a^{m + n} \ \ \ \ \ \ \ \ \dfrac{a^m}{a^n} = a^{m-n}$
Multiplying bases and exponents
Exponents and radicals
$a^{1 / n} = \sqrt[n]{a} \ \ \ \ \ \ \ \ \ \sqrt[n]{a^n} = a $
Zero exponent rule
$$a^0 = 1 \ \ \ \ \ (a ≠ 0)$$
OBS! You should not have to remember these by heart. The important thing is that you understand why these rules work. Then you can always recreate the rules by yourself if you happen to forget them.
Simplifying Exponents and Radicals
What is $\sqrt{2} \cdot \sqrt{3} $? The answer is simpler than you might first think. It's just $\sqrt{6} $. Taking the square root of both numbers and then multiplying them is the same as first multiplying them and then taking the square root of the result. This is the rule for multiplying bases $a^n \cdot b^n = (ab)^n$ in practice.
We can also show this in the following way:
$$ \sqrt{2} \cdot \sqrt{3} $$
$$ 2^{1 / 2} \cdot 3^{1/2} $$
$$ (2 \cdot3)^{1/2} $$
$$ \sqrt{2 \cdot3} $$
$$ \sqrt{6} $$
Exercise 1
Determine the value of the following expressions.
a) $\sqrt{3} \cdot \sqrt{27} $
b) $\sqrt{7} \cdot \sqrt{21} \cdot \sqrt{3} $
c) $\sqrt{ \dfrac{1}{8}} \cdot \sqrt{2} $
d) $\sqrt[3]{5} \cdot \sqrt[3]{25} $
a) $\sqrt{3} \cdot \sqrt{27} = $$ \ \sqrt{81} =$$ \ \mathbf{9} $
b) $\sqrt{7} \cdot \sqrt{21} \cdot \sqrt{3} =$$ \ \sqrt{ 7 \cdot21 \cdot3} =$$ \ \sqrt{21 \cdot21} =$$ \ \sqrt{21^2} =$$ \ \mathbf{21} $
c) $\sqrt{\dfrac{1}{8}} \cdot \sqrt{2} =$$ \ \sqrt{\dfrac{1}{4}} =$$ \ \mathbf{\dfrac{1}{2}} $
d) $\sqrt[3]{5} \cdot \sqrt[3]{25} =$$ \ \sqrt[3]{5^3} =$$ \ \mathbf{5} $
Exercise 2
Does $\sqrt{5} \ \sqrt[3]{7} = \sqrt{35} $?
Radicands can only be multiplied with each other if they are under the same radical. So $\sqrt{5} \ \sqrt[3]{7}$ does not equal $\sqrt{35}$. The expression $\sqrt{5} \sqrt[3]{7}$ can’t be simplified any further than what it already is.
When working with radicals, we prefer the radical number to be as small as possible. Take the $\sqrt{8} $ as an example. This can be simplified to:
$ 8^{1/2} \ = (2^2 \cdot 2)^{1/2} \ =$$ \ 2^{(2 \ \cdot \ 1/2)} \ \cdot \ 2^{1/2} \ =$$ \ 2 \sqrt{2} $
What we did here was to extract all the factors that don’t have to be below the radical sign. Instead of having $2^2 $ below the radical, we get $2$ outside the radical.
When simplifying radicals this way, it becomes easier to see which factors the expression is made of. By rewriting $\sqrt{8} $ to $2 \sqrt{2} $, we immediately see that the expression can be evenly divided by $2 $. This was not as obvious when the radical was written as $\sqrt{8} $.
The easiest way to simplify radicals is to first rewrite the number as a product of smaller factors. Take $\sqrt{96} $ as an example.
$ \sqrt{96} \ = (2^5 \cdot3)^{1/2} \ =$$ \ (2^4)^{1/2} \cdot (2 \cdot3)^{1/2} \ =$$ \ 2^2\cdot(2\cdot3)^{1/2}\ =$$ \ 4\sqrt{6}$
Taking the square root of a number has the same effect as dividing all the exponents of the factors that make up the number by $ 2 $. An easy method for simplifying square radicals is to divide all the factors’ exponents by $2$. The quotient of this division is what will go outside of the radical sign and the remainder is what will remain below the radical sign.
$2^5: \ \ \ \dfrac{5}{2}=2 + \text{remainder} \ 1 \ \ \Longrightarrow \ \ $$ 2^2=\mathbf{4} \ \text{outside radical,} \ \ \ 2^1=\mathbf{2} \ \text{below radical}$
$3^1: \ \ \ \dfrac{1}{2}=0 + \text{remainder} \ 1 \ \ \Longrightarrow \ \ $$ 3^0=\mathbf{1} \ \text{outside radical,} \ \ \ 3^1=\mathbf{3} \ \text{below radical}$
$(4 \cdot 1) \sqrt{2 \cdot 3}$
$\mathbf{4 \sqrt{6}}$
When working with cube radicals, we instead divide all the factors’ exponents by $3$ before sorting the quotient outside the radical and the remainder below the radical. With seventh root radicals, we divide by $ 7 $, and so on.
Exercise 3
Simplify the following expressions.
a) $ \sqrt{250}$ $(250 = 2 \cdot5^3) $
b) $ \sqrt[3]{864}$ $(864 = 2^5 \cdot3^3) $
c) $ \sqrt[4]{\dfrac{112}{1875}}$ $(112 = 2^4 \cdot 7 $, $ 1875 = 3 \cdot 5^4) $
a) $ \sqrt{250} \ = \ (2 \cdot5^3)^{1/2} \ =$$ \ (5^2)^{1/2} \cdot (2 \cdot5)^{1/2 } =$$ \ 5 \cdot \sqrt{2 \cdot 5} =$$ \ \mathbf{5 \sqrt{10}} $
b) $ \sqrt[3]{864} \ = \ (2^5 \cdot 3^3)^{1/3} \ = $$ \ (2^3 \cdot 3^3)^{1/3} \cdot (2^2)^{1/3} \ = $$ \ (2 \cdot 3) \cdot \sqrt[3]{2^2} \ = $$ \ \mathbf{6 \sqrt[3]{4}}$
c) $ \sqrt[4]{\dfrac{112}{1875}} \ \ = $$ \ \ \Big(\dfrac{2^4 \cdot 7}{3 \cdot 5^4} \Big)^{1 / 4} =$$ \ \ \Big(\dfrac{2^4}{5^4} \Big)^{1/4} \cdot \ \Big(\dfrac{7}{3} \Big)^{1 / 4} =$$ \ \ \ \dfrac{2}{5} \cdot \Big(\dfrac{7}{3} \Big)^{1/4} \ =$$ \ \ \mathbf{\dfrac{2}{5} \cdot \sqrt[4]{\dfrac{7}{3}}} $
Exercise 4
Simplify the following expression.
$$ \dfrac{8^{2/3} \cdot \ 2^{1/2} \cdot \ 9^{- 3/2}}{\sqrt{3} \ \cdot \ 2^{5/2 } \cdot \ 3^{- 3/2}} $$
We start by rewriting all factors to powers with bases $ 2 $ and $ 3 $.
$$ \dfrac{(2^3)^{2/3} \ \cdot \ 2^{1/2} \ \cdot \ (3^2)^{- 3/2}}{3^{1 / 2} \ \cdot \ 2^{5/2} \ \cdot \ 3^{- 3/2}} \ =$$
$$ \dfrac{2^{2} \ \cdot \ 2^{1/2} \ \cdot \ 3^{(-3)}}{3^{1/2} \ \cdot \ 2^{5/2} \cdot \ 3^{- 3/2}} \ =$$
$$ 2^{(2 + 1 / 2-5 / 2)} \ \ \cdot \ 3^{((-3) -1 / 2 + 3/2)} \ \ =$$
$$ 2^0 \cdot 3^{- 2} \ =$$
$$ \mathbf{\dfrac{1}{9}} $$
Exercise 5
Between which two integers does $\sqrt{72} $ lie?
$\sqrt{72} $ is the number that multiplied by itself becomes $72 $. By taking the square of different numbers, we see that $8^2 = 64 $ and $9^2 = 81 $, which is the same as $\sqrt{64} = 8 $ and $\sqrt{81} = 9 $. This means that $\sqrt{72} $ must lie between the numbers $\mathbf{8} $ and $\mathbf{9} $.
Negative Radicands
What happens if we take the square root of a negative number, such as $ \sqrt{-1} $? The value of this radical should be a number which multiplied itself equals $-1$. But obviously there is no such number, because a number squared is always positive.
Rather than stopping here, we can introduce a new number $\mathbf{i}$ so that $i = \sqrt{-1} $. This is a new class of numbers called complex numbers. Contrary to the real numbers, the complex numbers can’t be found anywhere on the number line. We will go through this topic in more detail in our lesson on complex numbers. For the moment, it’s enough that you know that the square root of a negative number can’t be found anywhere on the number line. As long as the root of the radical is even, such as $\sqrt[4]{-3}$ or $(-100)^{1/8}$, a negative radicand will always lead to a complex number.
There is no problem, however, with taking the an odd root of a negative number. For example, $ \sqrt[3]{-8} = -2 $ because $ (-2) \cdot (-2) \cdot (-2) = -8 $. We can extract the negative sign and place it outside the radical sign as long as the root of the radical is odd. For example, $ \sqrt[4]{-16} $ has no real solutions while $ \sqrt[5]{-1} = -1 $.
Exercise 6
Simplify the following expressions as much as possible:
a) $-\sqrt[3]{-54}$
b) $\dfrac{2}{5 \ \cdot \ 10^{-2}}$
c) $-(-8)^{-4/3}$
a) $-\sqrt[3]{-54} = -(-\sqrt[3]{2\cdot3^3} = $$ \ (2 \cdot 3^3)^{1/3} = $$ \ (3^3)^{1/3} \cdot 2^{1/3} = $$ \ \mathbf{3\sqrt[3]{2}}$
b) $\dfrac{2}{5 \cdot 10^{-2}} \ = $$ \ \dfrac{2}{5 \cdot \Big(\frac{1}{10^{2}} \Big)} \ = $$ \ \dfrac{2}{5 \cdot \frac{1}{100}} \ = $$ \dfrac{2 \cdot 100}{5} \ = $$ \mathbf{40}$
c) $-(-8)^{-4/3} \ = $$ \ -((-2)^3)^{-4/3} \ = $$ \ -(-2)^{(3 \ \cdot \ -4/3)} \ = $$ \ -(-2)^{-4} \ = \ -\dfrac{1}{(-2)^{4}} \ = $$ \ \mathbf{-\dfrac{1}{16}}$
Problems to Solve
Exercise 1
Determine the following expression and state your answer in the simplest possible form without exponents.
$$ \dfrac{(-2)^4}{(3^2 + 2^2)^2} $$
$$ \dfrac{(- 2)^4}{(3^2 + 2^2)^2} \ = $$
$$\dfrac{16}{(9 + 4)^2} \ =$$
$$\dfrac{16}{(13)^2} \ =$$
$$\mathbf{\dfrac{16}{169}}$$
Exercise 2
Determine $ (-3)^{- 2} + (- 2)^{- 1} + (- 1)^0 + 0^1 + 1^2 + 2^3 + 3^4 $.
If we determine each term separately and then add them, we get the following:
$$ (- 3)^{- 2} + (- 2)^{- 1} + (- 1)^0 + 0^1 + 1^2 + 2^3 + 3^4 \ =$$
$$ \dfrac{1}{9} + \dfrac{1}{- 2} + 1 + 0 + 1 + 8 + 81 \ =$$
$$ 91 + \dfrac{1}{9} - \dfrac{1}{2} \ =$$
$$ \mathbf{90 \dfrac{11}{18}} $$
Exercise 3
Simplify the expression $ \sqrt{8} \cdot \sqrt[3]{15} \cdot \sqrt{18} \cdot \sqrt[3]{9} \cdot \sqrt[3]{25}$.
We multiply the numbers below the square radicals and the numbers below the cube radicals separately. This gives us:
$$ \sqrt{8 \cdot18} \ \cdot \ \sqrt[3]{15 \cdot9 \cdot25} \ =$$
$$ \sqrt{2^3 \cdot (2 \cdot3^2)} \ \cdot \ \sqrt[3]{(3 \cdot5) \cdot3^2 \cdot5^2} \ =$$
$$ \sqrt{2^4 \cdot3^2} \ \cdot \ \sqrt[3]{3^3 \cdot5^3} \ =$$
$$ (2^2 \cdot 3) \cdot (3 \cdot5) \ =$$
$$ \mathbf{180} $$
Exercise 4
Rank the expressions below in order of size with the smallest number first and the largest number last.
$$ (-2)^{- 3}, \ \ \sqrt{2}, \ \ 8^{2/3}, \ - (5^{- 1}), \ \ 1000^0 $$
We can determine the five numbers as follows:
- $(- 2)^{- 3} = \dfrac{1}{{- 2}^3} = \dfrac{1}{8} $
- $1 < \sqrt{2}
Exercise 5
Determine $ 16^{- (2^{- 2})} $.
We start with the exponents that are highest in the order first. This gives us:
$$ 16^{- (2^{- 2})} \ = $$
$$16^{- (1/4)} \ = $$ \
$$\dfrac{1}{16^{1/4}} \ =$$
$$\mathbf{\dfrac{1}{ 2}} $$
Exercise 6
What is $ ((((25^3 \cdot 4^5 - 6 \cdot3 + 2)^2)^5)^0)^7 $?
Using the rule for multiplying exponents we get:
$$ (25^3 \cdot 4^5 - 6 \cdot3 + 2)^{(2 \cdot5 \cdot0 \cdot7)} \ \ =$$
$$ (25^3 \cdot 4^5 - 6 \cdot3 + 2)^{0} $$
Any number raised to $0 $ always becomes $1 $ (except $0^0 $), so the answer is simply $\mathbf{1} $.
Exercise 7
What is $ a $ if $ \sqrt{3+ \sqrt{1+ \sqrt{a}}} \ = \ 2 $?
To get rid of a square root we raise it to the power of $2 $. If we continue to square both sides of the equation, we can systematically remove all radicals.
$$ \sqrt{3+ \sqrt{1+ \sqrt{a}}} \ = \ 2 $$
$$ \bigg(\sqrt{3+ \sqrt{1+ \sqrt{a}}} \ \bigg)^2 = \ 2^2 $$
$$ 3 + \sqrt{1+ \sqrt{a}} = \ 4 $$
$$ \sqrt{1+ \sqrt{a}} = \ 1 $$
$$ \Big(\sqrt{1+ \sqrt{a}} \ \Big)^2 = \ 1^2 $$
$$ 1 + \sqrt{a} = \ 1 $$
$$ \sqrt{a} = \ 0 $$
$$ \mathbf{a = 0} $$
Exercise 8
Simplify
$$\dfrac{(x^8)(12y^{-5/2}) \Big( \dfrac{3}{z^{-7}} \Big) }{\Big( \dfrac{x^{-4}}{5y^3} \Big)^2 9(z^{(1/2)^{-2}})}$$
This is a very complicated expression, but if we take it one step at a time we can solve it! We begin to simplify the factors in the denominator first.
$$ \dfrac{(x^8) (12y^{- 5/2}) \Big(\dfrac{3}{z^{- 7}} \Big)}{\Big(\dfrac{x^{ -4}}{5y^3} \Big)^2 9 (z^{(1/2)^{- 2}})} \ =$$
$$ \dfrac{(x^8) (12y^{- 5/2}) \Big(\dfrac{3}{z^{- 7}} \Big)}{\dfrac{(x^{- 4 })^2}{{(5y^3)^2}} \ 9 (z^{(\frac{1}{1/2})^{2}})} \ =$$
$$ \dfrac{(x^8) (12y^{- 5/2}) \Big(\dfrac{3}{z^{- 7}} \Big)}{\dfrac{x^{- 8} }{{25y^6}} \ 9 (z^{(2)^{2}})} \ =$$
$$ \dfrac{(x^8) (12y^{- 5/2}) \Big(\dfrac{3}{z^{- 7}} \Big)}{(x^{- 8}) \Big(\dfrac{y^{- 6}}{25} \Big) \ 9 (z^{4})} \ =$$
$$ \dfrac{25 (x^8) (12y^{- 5/2}) \Big(\dfrac{3}{z^{- 7}} \Big)}{9 (x^{- 8} ) (y^{- 6}) (z^{4})} \ = $$
Let us now move on to simplifying the numerator.
$$ \dfrac{25 (x^8) (12y^{- 5/2}) (3z^{- (- 7)})}{9 (x^{- 8}) (y^{- 6} ) (z^{4})} \ = $$
$$ \dfrac{25 (x^8) (12y^{- 5/2}) (3z^{7})}{9 (x^{- 8}) (y^{- 6}) (z^{4})} \ = $$
$$ \dfrac{(25) (12) (3) (x^8) (y^{- 5/2}) (z^{7})}{9 (x^{- 8}) (y^{-6}) (z^{4})} \ = $$
$$ \dfrac{(25) (4) (1) (x^8) (y^{- 5/2}) (z^{7})}{(x^{- 8}) (y^{ -6}) (z^{4})} \ =$$
$$ \dfrac{(100) (x^8) (y^{- 5/2}) (z^{7})}{(x^{- 8}) (y^{- 6}) (z^{4})} \ =$$
We can now rewrite the fraction as a single product by changing the signs of the exponents in the numerator.
$$ (100) (x^8) (y^{- 5/2}) (z^{7}) (x^{8}) (y^{6}) (z^{- 4}) \ =$$
$$ 100 (x^{8 + 8}) (y^{- 5/2 + 6}) (z^{7-4}) \ =$$
So the final answer is:
$$ \mathbf{100 (x^{16}) (y^{7/2}) (z^{3})} $$
Exercise 9
Find \(x\) if \(3^{27^x} = 27^{3^x}\).
Note that $27 = 3^3 $. By rewriting $27$ in exponential form we get:
$$ 3^{27^x} = 27^{3^x} $$
$$ 3^{(3^3)^x} = (3^3)^{3^x} $$
$$ 3^{3^{3 \ \cdot \ x}} = 3^{3 \ \cdot \ {3^x}} $$
Both the left and the right-hand side consist of a power with the base $3 $. Since both powers in the equation must be equal, it follows that their exponents must be equal. We can thus remove the base $3 $ from both sides of the equation:
$$3^{3 \ \cdot \ x} = 3 \cdot {3^x} $$
$$ 3^{3x} = {3^{x + 1}} $$
Here too we can remove the base $3 $ from both sides in the equation:
$$ 3x = x + 1 $$
$$ \mathbf{x = \dfrac{1}{2}} $$