Order of Operations - Exponents and Roots

πŸ†Practice powers and roots

Order of Operations - Exponents and Roots

The second step in the order of operations is - exponents and roots!

2 English Order of Operations Exponents

Immediately after dealing with parentheses, we move on to exponents and roots!
Pay attention - even within parentheses, it's very important to maintain the correct order of operations!
Exponent - multiply the base by itself the number of times shown in the exponent (the small number on the top right).
Root - half power - which positive number when multiplied by itself will give the number written under the root.

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Test yourself on powers and roots!

einstein

\( 6+\sqrt{64}-4= \)

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Order of Operations - Exponents and Roots

Reminder - How to solve exponents?

An exponent is the requirement for a number to be multiplied by itself several times.
In other words, when we see a number with a certain exponent, we know that we need to multiply the number by itself several times to get the actual number.
The base of the exponent will be the number itself and the exponent will be the power number - the small one on the top right.
For example:
53=5^3=
The number 55 is raised to the power of 33 so we need to multiply the number 55 by itself 33 times. We get:
5β‹…5β‹…5=1255\cdot5\cdot5=125
Remember - when a number is raised to the power of 00 = the result will always be 11.

Reminder - How to solve a root?

A root equals to the power of 0.50.5
The result of a root will always be positive or 00 and never negative.
When we encounter a root of a number, it's like asking ourselves - which number raised to the power of 22 would give us the number inside the root.
For example:
16=42\sqrt{16}=4^2
When we see the root of 1616 we ask ourselves, which number squared equals 1616?
The answer is 44.

It's important that you know the root formulas:

Root of a product

(aβ‹…b)=aβ‹…b\sqrt{(a\cdot b)}=\sqrt{a}\cdot\sqrt{b}

Square root of a quotient

ab=ab\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}

square root of a square root

amn=anβ‹…m\sqrt[n]{\sqrt[m]{a}}=\sqrt[n\cdot m]{a}

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Exponents and Roots in Order of Operations

When it comes to order of operations, right after parentheses we deal with exponents and roots in the expression!
The truth is? It doesn't matter if we start from left to right. At this stage of handling exponents and roots, we are only changing the expression cosmetically, and as long as we deal with all exponents and roots, their order doesn't matter.
Let's practice:

Here is an exercise –
42βˆ’25+(32+12300)=4^2-\sqrt{25}+(3^2+1230^0 )=

Solution:
We learned that the first operation in the order of operations is dealing with parentheses, so we'll handle the parentheses first.
Notice that inside the parentheses there are numbers with exponents, so we'll solve them first –
32=93^2=9
12300=11230^0=1
Let's write the results we got inside the parentheses in the original expression and we get:
42βˆ’25+(9+1)=4^2-\sqrt{25}+(9+1 )=
Let's continue solving –
Notice, have we finished dealing with parentheses? Not really. We'll solve the expression inside the parentheses and only then can we continue with the other numbers.
Let's solve the expression inside the parentheses:
9+1=109+1=10
And now let's write the original expression without parentheses:
42βˆ’25+10=4^2-\sqrt{25}+10=
Now let's continue solving – the next step is solving exponents and roots.
42=164^2=16
25=5\sqrt{25}=5
Let's write the results in the original expression and we get:
16βˆ’5+10=16-5+10=
Now this is a really easy expression to solve. Let's solve it and we get:
​​​​​​​16βˆ’5+10=21​​​​​​​16-5+10=21

Let's practice another exercise!
40βˆ’(3β‹…22βˆ’4)+(25βˆ’2)β‹…2β‹…3βˆ’(22:40)=40-(3\cdot2^2-4)+(\sqrt{25}-2)\cdot2\cdot3-(2^2:4^0 )=

Solution:
We know this exercise is a bit long... but don't worry. Let's solve it slowly, step by step:
Let's start with the first step in order of operations - parentheses!
The first parentheses from left to right are:
(3β‹…22βˆ’4)=(3\cdot2^2-4)=
We know that the next step right after is exponents and roots, so we'll deal with the exponent inside the parentheses:
(3β‹…4βˆ’4)=(3\cdot4-4)=
Now we'll solve the multiplication and then the subtraction and get:
(3β‹…4βˆ’4)=8(3\cdot4-4)=8
We'll write the result in the original exercise and continue:
40βˆ’8+(25βˆ’2)β‹…2β‹…3βˆ’(22:40)=40-8+(\sqrt{25}-2)\cdot2\cdot3-(2^2:4^0 )=
Let's move to the second parentheses:
(25βˆ’2)=(\sqrt{25}-2)=
Let's deal with the square root
25=5\sqrt{25}=5
And we get:
(25βˆ’2)=(5βˆ’2)=3(\sqrt{25}-2)=(5-2)=3

Let's write the result in the original exercise and continue solving.
40βˆ’8+3β‹…2β‹…3βˆ’(22:40)=40-8+3\cdot2\cdot3-(2^2:4^0 )=
Now let's continue solving the parentheses:
(22:40)=(2^2:4^0 )=
22=42^2=4
40=14^0=1
Let's write the results in the parentheses and solve:
(4:1)=4(4:1)=4
Great! Now let's write the result in the original exercise instead of the parentheses and we get:
40βˆ’8+3β‹…2β‹…3βˆ’4=40-8+3\cdot2\cdot3-4=
Now that we've gotten rid of all parentheses, roots, and exponents, let's continue to the third step - multiplication and division from left to right:
3β‹…2β‹…3=183\cdot2\cdot3=18
Let's write the result in the original exercise:
40βˆ’8+18βˆ’4=40-8+18-4=
We're done with multiplication and division and can move on to addition and subtraction from left to right. We get:
32+18βˆ’4=32+18-4=
50βˆ’4=4650-4=46
The result is 4646!

Do you know what the answer is?

Examples with solutions for Powers and Roots

Exercise #1

6+64βˆ’4= 6+\sqrt{64}-4=

Video Solution

Step-by-Step Solution

To solve the expression 6+64βˆ’4= 6+\sqrt{64}-4= , we need to follow the order of operations (PEMDAS/BODMAS):


  • P: Parentheses (or Brackets)
  • E: Exponents (or Orders, i.e., powers and roots, etc.)
  • MD: Multiplication and Division (left-to-right)
  • AS: Addition and Subtraction (left-to-right)

In this expression, we first need to evaluate the square root since it falls under the exponent category:


64=8 \sqrt{64} = 8


Next, we substitute the computed value back into the expression:


6+8βˆ’4 6+8-4


We then perform the addition and subtraction from left to right:


6+8=14 6+8 = 14


14βˆ’4=10 14-4 = 10


Thus, the final answer is:


10 10

Answer

10

Exercise #2

10:2βˆ’22= 10:2-2^2=

Video Solution

Step-by-Step Solution

The given mathematical expression is 10:2βˆ’22 10:2-2^2 .

According to the order of operations (often remembered by the acronym PEMDAS/BODMAS), we perform calculations in the following sequence:

  • Parentheses/Brackets
  • Exponents/Orders (i.e., powers and roots)
  • Multiplication and Division (from left to right)
  • Addition and Subtraction (from left to right)

In this expression, there are no parentheses, but there is an exponent: 222^2. We calculate the exponent first:

22=42^2 = 4

Substituting back into the expression, we have:

10:2βˆ’4 10:2-4

Next, we perform the division from left to right. Here, ":" is interpreted as division:

10Γ·2=5 10 \div 2 = 5

Now, substitute this back into the expression:

5βˆ’4 5 - 4

The final step is to perform the subtraction:

5βˆ’4=1 5 - 4 = 1

Therefore, the answer is 1 1 .

Answer

1

Exercise #3

3Γ—3+32=Β ? 3\times3+3^2=\text{ ?}

Video Solution

Step-by-Step Solution

First we need to remind ourselves of the order of operations:

  1. Parentheses

  2. Exponents and Roots

  3. Multiplication and Division

  4. Addition and Subtraction

There are no parentheses in this problem, therefore we will start with exponents:

3 * 3 + 3Β² =

3 * 3 + 9 =

Let's continue to the next stepβ€”multiplication operations:

3 * 3 + 9 =

9 + 9 =

Finally, we are left with a simple addition exercise:

9 + 9 = 18

Answer

18

Exercise #4

8βˆ’32:3= 8-3^2:3=

Video Solution

Step-by-Step Solution

Let's solve the expression step by step using the order of operations, often remembered by the acronym PEMDAS, which stands for Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right).

The given expression is: 8βˆ’32:3= 8-3^2:3=

Step 1: Evaluate Exponents
The expression has an exponent, which we need to evaluate first. The exponent is 323^2.
Calculate 323^2 which equals 99.
Now the expression becomes: 8βˆ’9:3 8 - 9 : 3

Step 2: Division
Next, perform the division operation. Here we divide 99 by 33.
Calculate 9:39 : 3 which equals 33.
Now the expression becomes: 8βˆ’3 8 - 3

Step 3: Subtraction
Finally, perform the subtraction.
Calculate 8βˆ’38 - 3 which equals 55.

Therefore, the solution to the expression 8βˆ’32:38-3^2:3 is 55.

Answer

5 5

Exercise #5

4+2+52= 4+2+5^2=

Video Solution

Step-by-Step Solution

To solve the expression 4+2+52 4 + 2 + 5^2 , we need to follow the order of operations, often remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction).

  • Step 1: Calculate Exponents
    In the expression we have an exponent: 525^2. This means 5 is raised to the power of 2. We calculate this first:
    52=255^2 = 25.

  • Step 2: Perform Addition
    Now, substitute the calculated value back into the expression:
    4+2+254 + 2 + 25.
    Perform the additions from left to right:
    4+2=64 + 2 = 6
    Finally add the result to 25:
    6+25=316 + 25 = 31.

Therefore, the final answer is 3131.

Answer

31

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