Application of Laws of Exponents and Scientific Notations

APPLICATION OF LAWS OF EXPONENTS AND SCIENTIFIC NOTATIONS

INTRODUCTION

The objective for this lesson on Application of Laws of Exponents and Scientific Notation is, the student will use properties of exponents in order to solve problems involving operations with scientific notation.

The skills students should have in order to help them in this lesson include scientific notation and standard form.

We will have three essential questions that will be guiding our lesson. Number 1, how do we add and subtract numbers written in scientific notation with the same exponents? Number 2, how do we multiply or divide numbers written in scientific notation? And number 3, what does “E” followed by a number represent on a calculator?

We will begin by completing the warm-up using our knowledge of scientific notation and standard form to translate between two forms to prepare for application of laws of exponents and scientific notation in this lesson.

SOLVE PROBLEM INTRODUCTION

The SOLVE problem for this lesson is, in a recent study, results explained that approximately one point five times ten to the third power bottles of water were consumed each second. Given that there are three point one five three six times ten to the seventh power seconds in a year, how many bottles of water are expected to be consumer in a year?

We will begin by Studying the Problem. First we need to identify where the question is located within the problem and underline the question. How many bottles of water are expected to be consumed in a year?

Now that we have identified the question, we want to put this question in our own words in the form of a statement. This problem is asking me to find the number of bottles of water consumed in a year.

During this lesson we will learn how to solve problems by applying laws of exponents and scientific notation. We will use this knowledge to complete this SOLVE problem at the end of the lesson.

ADDITION AND SUBTRACTION WITH THE SAME EXPONENTS

We are going to explore and identify patterns for adding and subtracting numbers in scientific notation that have the same exponents.

Let’s take a look at Problem one seen here. The problem is seven point five two times ten to the third power plus two point one nine times ten to the third power.

What does Problem one ask us to find? It asks us to find the sum of the two numbers written in scientific notation. What are some possible strategies for adding the two values? Think about that for just a moment.

Now using the graphic organizer that you have in front of you, explain one way that we can solve this problem. We could translate the values to standard form to solve the problem.

What is the first number translated to standard form? The first number in scientific notation is seven point five two times ten to the third power. Remember that in a previous lesson we discovered that the exponent on the ten tells us how many places to move the decimal point to the right or the left in the value of seven point five two. Since the exponent is positive we will move to the right. The exponent is the number three, which tells us to move the decimal point three places. When we move the decimal point in seven point five two three places to the right we get seven thousand five hundred twenty.

What is the second number translated to standard form? Let’s follow the same steps, two point one nine times ten to the third power. Theexponent of three on the ten tells us that we will move the decimal point in two point one nine three places to the right. This gives us the value of two thousand one hundred ninety, when we put the number in scientific notation into standard form.

Now what is the sum of the two numbers written in standard form? Seven thousand five hundred twenty plus two thousand one hundred ninety equals nine thousand seven hundred ten.

Record this under the “Solution in Standard Form” in your graphic organizer.

Now if you convert nine thousand seven hundred ten back to scientific notation, what is the solution? Remember that when we write a number in scientific notation we need to write the number between one and ten, and then multiply it by a power of ten. In scientific notation nine thousand seven hundred ten is nine point seven one times ten to the third power. Record this information under “Solution in Scientific Notation” in your graphic organizer.

Now let’s make a few observations about this problem.

What was the exponent of the first number in the original problem? It was three.

And what was the exponent of the second number in the original problem? It was also three.

What was the exponent of the solution? Again the exponent was three.

Now what was the coefficient of the first number? In the original problem the coefficient of the first number is seven point five two.

And what was the coefficient of the second number in the original problem? It was two point one nine.

And what was the coefficient of the solution? Nine point seven one.

Let’s take a look at another problem together. Problem two is six point two eight times ten to the negative fourth power plus three point three four times ten to the negative fourth power.

What does Problem two ask us to find? It asks us to find the sum of the two numbers written in scientific notation.

How are the values in Problem two different than the values in Problem one? The exponents are negative, which means we’re working with small numbers.

How did we solve Problem one? We translated the values to standard form? Let’s do that for this problem as well.

What is the first number translated to standard form? Remember that in previous lesson we discovered that the exponent on the ten tells us how many places to move the decimal point to the right or left in the value.

For this first number six point two eight times ten to the negative fourth power, we will need to move the decimal point four places as indicated by the exponent. Since the exponent is negative we will need to move the decimal four places to the left. When we do so, this number in standard form is zero point zero zero zero six two eight.

Now what is the second number translated to standard form? Again we need to look at the exponent on the ten. It is also negative four, which means we will need to move the decimal point in the value four places to the left. When we do so, this number written in standard form is zero point zero zero zero three three four.

What is the sum of the two numbers written in standard form? Zero point zero zero zero six two eight plus zero point zero zero zero three three four equals zero point zero zero zero nine six two.

Record this information under “Solution in Standard Form” in the graphic organizer on your page.

Now if you convert zero point zero zero zero nine six two back to scientific notation, what is the solution? Remember that when we write a number in terms of scientific notation it needs to be a number between one and ten, which is multiplied by a power of ten. When we convert this number back to scientific notation we get nine point six two times ten to the negative fourth power.

Record this information under “Solution in Scientific Notation” in the graphic organizer on your page.

Now let’s make some observations about this problem.

What was the exponent of the first number in Problem two? It was negative four.

And what was the exponent of the second number in Problem two? It was also negative four.

And what was the exponent of the solution? Again, the exponent was negative four.

What was the coefficient of the first number? It was six point two eight.

And what was the coefficient of the second number? It was three point three four.

And what was the coefficient of the solution? Nine point six two.

Now let’s take a look at a third problem together.

Problem three is four point two five times ten to the fifth power minus three point one six times ten to the fifth power.

What does Problem three ask us to find? It asks us to find the difference of the two numbers written in scientific notation.

Using the graphic organizer, explain one way that we can solve this problem. We can translate the values to standard form. Let’s do so now.

What is the first number translated to standard form? Four point two five times ten to the fifth power, when written in standard form is four hundred twenty five thousand.

And what is the second number translated to standard form? Three point one six times ten to the fifth power, when written in standard form is three hundred sixteen thousand.

What is the difference of the two numbers written in standard form? Four hundred twenty five thousand minus three hundred sixteen thousand equals one hundred nine thousand. Record this information under the “Solution in Standard Form” box on your graphic organizer.

If you convert one hundred nine thousand back to scientific notation, what is the solution? Remember that scientific notation is a number between one and ten, which is multiplied by a power of ten. One hundred nine thousand in scientific notation is written as one point zero nine times ten to the fifth power. Record this information under “Solution in Scientific Notation” on your graphic organizer.

Let’s make some observations about this problem as well.

What was the exponent of the first number in Problem three? It was five.

And what was the exponent of the second number in Problem three? Also five.

What was the exponent of the solution for Problem three? Again, it was five.

What was the coefficient of the first number in Problem three? It was four point two five.

And what was the coefficient of the second number in Problem three? It was three point one six.

And what was the coefficient of the solution? It was one point zero nine.

Now let’s take a look at Problem four.

Problem four is, nine point one two times ten to the negative second power minus five point zero six times ten to the negative second power.

What does Problem four ask us to find? It is asking us to find the difference of the two numbers written in scientific notation.

How is this problem different from Problem three? The exponents in this problem are negative.

Using the graphic organizer in front of you, explain one way that we can solve this problem. We can translate the values to standard form to solve. Let’s do that now.

What is the first number translated to standard form? The first number in scientific notation is nine point one two times ten to the negative second power. Since the exponent on the ten is a negative two we need to move the decimal point two places to the left in the value nine point one two. Written in standard form this number is zero point zero nine one two.

What is the second number translated to standard form? The second number in scientific notation is five point zero six times ten to the negative second power. Since the exponent on the ten is a negative two this tells us again that we need to move the decimal point two places to the left in the value five point zero six. When we do so this number written in standard form is zero point zero five zero six.

What is the difference of the two numbers written in standard form? Zero point zero nine one two minus zero point zero five zero six equals zero point zero four zero six. Record this under the “Solution in Standard Form” on your graphic organizer.

If you convert zero point zero four zero six back to scientific notation, what is the solution? Remember when we write a number in scientific notation we need to write a number between one and ten, which is multiplied by a power of ten. Zero point zero four zero six written in scientific notation is four point zero six times ten to the negative second power. Record this information under “Solution in Scientific Notation” on your graphic organizer.

Now let’s make so observations together about this problem.

What was the exponent of the first number in Problem four? It was negative two.

And what was the exponent of the second number in Problem four? Also negative two.

What was the exponent of the solution for Problem four? Again negative two.

What was the coefficient of the first number in Problem four? Nine point one two.

And what was the coefficient of the second number in Problem four? Five point zero six.

Finally what was the coefficient of the solution? Four point zero six.

Now let’s summarize the work that we’ve done together. Take a look at the exponents in the bottom row of each problem. Explain what you notice about the exponents when adding or subtracting numbers in scientific notation. When we add or subtract the numbers, the exponent of the solution stayed the same.

What generalizations can you make regarding the coefficients of the numbers in scientific notation? With addition, we add the decimals of the coefficients to get the coefficient of the solution. With subtraction, we subtract the decimals of the coefficients to get the coefficient of the solution.

So to summarize, if the exponents of the numbers are the same, we can add or subtract the coefficient and keep the exponent.

ADJUSTING SOLUTIONS WITH ADDITION AND SUBTRACTION

Let’s take a look at Problem five together.

The problem is six point five one times ten to the third power plus four point two one times ten to the third power.

Using what we discovered in the previous section, that if the exponents of the numbers are the same we can add or subtract the coefficient and keep the exponents.

Let’s solve Problem five together. We will add the coefficients together. Six point five one plus four point two one which is ten point seven two. And we will keep the exponents the same. So the solution is ten point seven two times ten to the third power.

Let’s also solve Problem six using what we discovered in the previous section.

This time the problem is seven point nine seven times ten to the negative fourth power minus seven point nine three times ten to the negative fourth power.

This time we need to subtract the coefficients. Seven point nine seven minus seven point nine three equals zero point zero four. We will keep the exponents the same.

Our solution is zero point zero four times ten to the negative fourth power.

Look at these solutions for Problems five and six. What is different about these solutions compared to the solutions to Problems one through four? The coefficients of the solutions are not written in true scientific notation because Problem five has two digits to the left of the decimal point and Problem six does not have a value between one and ten to the left of the decimal point.

To correctly write a number in scientific notation, we should have one digit to the left of the decimal point.

Looking at Problem five and the solution that we came up with, use the space in the last row to convert to standard form, finding the solution and then converting back to scientific notation.

What is the solution to Problem five? If we change the first number to standard form, six point five one times ten to the third power is equal to six thousand five hundred ten. And four point two one times ten to the third power is equal to four thousand two hundred ten. We will add six thousand five hundred ten plus four thousand two hundred ten, which equals ten thousand seven hundred twenty. Written in scientific notation ten thousand seven hundred twenty is one point zero seven two times ten to the fourth power.

Now looking at Problem six, use the space in the last row to convert to standard form, finding the solution, and then converting back to scientific notation. What is the solution to Problem six? Converting both of the numbers in the original problem to standard form we will have zero point zero zero zero seven nine seven minus zero point zero zero zero seven nine three. When we subtract we get zero point zero zero zero zero zero four. We can write this solution in scientific notation as four point zero zero times ten to the negative sixth power.

Looking at the solution to Problem five, what did you notice about the solution to when you used standard form? We converted the standard form solution to scientific notation and the decimal point moved to the left one digit and the exponent increased by one.