problem solving involving scientific notation

Module 7: Exponents

Problem solving with scientific notation, learning outcome.

Solve application problems involving scientific notation

Molecule of water with one oxygen bonded to two hydrogen.

A water molecule.

Solve Application Problems

Learning rules for exponents seems pointless without context, so let us explore some examples of using scientific notation that involve real problems. First, let us look at an example of how scientific notation can be used to describe real measurements.

Think About It

Match each length in the table with the appropriate number of meters described in scientific notation below. Write your ideas in the textboxes provided before you look at the solution.

Several red blood cells.

One of the most important parts of solving a “real-world” problem is translating the words into appropriate mathematical terms and recognizing when a well known formula may help. Here is an example that requires you to find the density of a cell given its mass and volume. Cells are not visible to the naked eye, so their measurements, as described with scientific notation, involve negative exponents.

Human cells come in a wide variety of shapes and sizes. The mass of an average human cell is about [latex]2\times10^{-11}[/latex] grams. [1] Red blood cells are one of the smallest types of cells [2] , clocking in at a volume of approximately [latex]10^{-6}\text{ meters }^3[/latex]. [3] Biologists have recently discovered how to use the density of some types of cells to indicate the presence of disorders such as sickle cell anemia or leukemia. [4] Density is calculated as [latex]\frac{\text{ mass }}{\text{ volume }}[/latex]. Calculate the density of an average human cell.

Read and Understand: We are given an average cellular mass and volume as well as the formula for density. We are looking for the density of an average human cell.

Define and Translate: [latex]m=\text{mass}=2\times10^{-11}[/latex], [latex]v=\text{volume}=10^{-6}\text{ meters}^3[/latex], [latex]\text{density}=\frac{\text{ mass }}{\text{ volume }}[/latex]

Write and Solve: Use the quotient rule to simplify the ratio.

[latex]\begin{array}{c}\text{ density }=\frac{2\times10^{-11}\text{ grams }}{10^{-6}\text{ meter }^3}\\\text{ }\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\times10^{-11-\left(-6\right)}\frac{\text{ grams }}{\text{ meter }^3}\\\text{ }\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\times10^{-5}\frac{\text{ grams }}{\text{ meter }^3}\\\end{array}[/latex]

If scientists know the density of healthy cells, they can compare the density of a sick person’s cells to that to rule out or test for disorders or diseases that may affect cellular density.

The average density of a human cell is [latex]2\times10^{-5}\frac{\text{ grams }}{\text{ meter }^3}[/latex]

The following video provides an example of how to find the number of operations a computer can perform in a very short amount of time.

Earth in the foreground, sun in the background, light beams traveling from the sun to the earth.

Light traveling from the sun to the earth.

In the next example, you will use another well known formula, [latex]d=r\cdot{t}[/latex], to find how long it takes light to travel from the sun to Earth. Unlike the previous example, the distance between the earth and the sun is massive, so the numbers you will work with have positive exponents.

The speed of light is [latex]3\times10^{8}\frac{\text{ meters }}{\text{ second }}[/latex]. If the sun is [latex]1.5\times10^{11}[/latex] meters from earth, how many seconds does it take for sunlight to reach the earth? Write your answer in scientific notation.

Read and Understand: We are looking for how long—an amount of time. We are given a rate which has units of meters per second and a distance in meters. This is a [latex]d=r\cdot{t}[/latex] problem.

Define and Translate:

[latex]\begin{array}{l}d=1.5\times10^{11}\\r=3\times10^{8}\frac{\text{ meters }}{\text{ second }}\\t=\text{ ? }\end{array}[/latex]

Write and Solve: Substitute the values we are given into the [latex]d=r\cdot{t}[/latex] equation. We will work without units to make it easier. Often, scientists will work with units to make sure they have made correct calculations.

[latex]\begin{array}{c}d=r\cdot{t}\\1.5\times10^{11}=3\times10^{8}\cdot{t}\end{array}[/latex]

Divide both sides of the equation by [latex]3\times10^{8}[/latex] to isolate t.

[latex]\begin{array}{c}\frac{1.5\times10^{11}}{3\times10^{8}}=\frac{3\times10^{8}}{3\times10^{8}}\cdot{t}\end{array}[/latex]

On the left side, you will need to use the quotient rule of exponents to simplify, and on the right, you are left with t.

[latex]\begin{array}{c}\text{ }\\\left(\frac{1.5}{3}\right)\times\left(\frac{10^{11}}{10^{8}}\right)=t\\\text{ }\\\left(0.5\right)\times\left(10^{11-8}\right)=t\\0.5\times10^3=t\end{array}[/latex]

This answer is not in scientific notation, so we will move the decimal to the right, which means we need to subtract one factor of [latex]10[/latex].

[latex]0.5\times10^3=5.0\times10^2=t[/latex]

The time it takes light to travel from the sun to Earth is [latex]5.0\times10^2[/latex] seconds, or in standard notation, [latex]500[/latex] seconds. That is not bad considering how far it has to travel!

Scientific notation was developed to assist mathematicians, scientists, and others when expressing and working with very large and very small numbers. Scientific notation follows a very specific format in which a number is expressed as the product of a number greater than or equal to one and less than ten times a power of [latex]10[/latex]. The format is written [latex]a\times10^{n}[/latex], where [latex]1\leq{a}<10[/latex] and n is an integer. To multiply or divide numbers in scientific notation, you can use the commutative and associative properties to group the exponential terms together and apply the rules of exponents.

Orders of magnitude (mass). (n.d.). Retrieved May 26, 2016, from https://en.wikipedia.org/wiki/Orders_of_magnitude_(mass) ↵
How Big is a Human Cell? ↵
How big is a human cell? - Weizmann Institute of Science. (n.d.). Retrieved May [latex]26, 2016[/latex], from http://www.weizmann.ac.il/plants/Milo/images/humanCellSize120116Clean.pdf ↵
Grover, W. H., Bryan, A. K., Diez-Silva, M., Suresh, S., Higgins, J. M., & Manalis, S. R. (2011). Measuring single-cell density. Proceedings of the National Academy of Sciences, 108(27), 10992-10996. doi:10.1073/pnas.1104651108 ↵
Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
Application of Scientific Notation - Quotient 1 (Number of Times Around the Earth). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/san2avgwu6k . License : CC BY: Attribution
Application of Scientific Notation - Quotient 2 (Time for Computer Operations). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/Cbm6ejEbu-o . License : CC BY: Attribution
Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology and Education. Located at : http://nrocnetwork.org/dm-opentext . License : CC BY: Attribution

The Math You Need, When You Need It

math tutorials for students majoring in the earth sciences

Scientific Notation - Practice Problems

Solving earth science problems with scientific notation, × div[id^='image-'] {position:static}div[id^='image-'] div.hover{position:static} introductory problems.

These problems cover the fundamentals of writing scientific notation and using it to understand relative size of values and scientific prefixes.

Problem 1: The distance to the moon is 238,900 miles. Write this value in scientific notation.

Problem 2: One mile is 1609.34 meters. What is the distance to the moon in meters using scientific notation?

`1609.34 m/(mi) xx 238","900 mi` = 384,400,000 m

Notice in the above unit conversion the 'mi' units cancel each other out because 'mi' is in the denominator for the first term and the numerator for the second term

Problem 4: The atomic radius of a magnesium atom is approximately 1.6 angstroms, which is equal to 1.6 x 10 -10 meters (m). How do you write this length in standard form?

0.00000000016 m

Fissure A = 40,0000 m Fissure B = 5,0000 m

This shows fissure A is larger (by almost 10 times!). The shortcut to answer a question like this is to look at the exponent. If both coefficients are between 1-10, then the value with the larger exponent is the larger number.

Problem 6: The amount of carbon in the atmosphere is 750 petagrams (pg). One petagram equals 1 x 10 15 grams (g). Write out the amount of carbon in the atmosphere in (i) scientific notation and (ii) standard decimal format.

The exponent is a positive number, so the decimal will move to the right in the next step.

750,000,000,000,000,000 g

Advanced Problems

Scientific notation is used in solving these earth and space science problems and they are provided to you as an example. Be forewarned that these problems move beyond this module and require some facility with unit conversions, rearranging equations, and algebraic rules for multiplying and dividing exponents. If you can solve these, you've mastered scientific notation!

Problem 7: Calculate the volume of water (in cubic meters and in liters) falling on a 10,000 km 2 watershed from 5 cm of rainfall.

`10,000 km^2 = 1 xx 10^4 km^2`

5 cm of rainfall = `5 xx 10^0 cm`

Let's start with meters as the common unit and convert to liters later. There are 1 x 10 3 m in a km and area is km x km (km 2 ), therefore you need to convert from km to m twice:

`1 xx 10^3 m/(km) * 1 xx 10^3 m/(km) = 1 xx 10^6 m^2/(km)^2` `1 xx 10 m^2/(km)^2 * 1 xx 10^4 km^2 = 1 xx 10^10 m^2` for the area of the watershed.

For the amount of rainfall, you should convert from centimeters to meters:

`5 cm * (1 m)/(100 cm)= 5 xx 10^-2 m`

`V = A * d`

When multiplying terms with exponents, you can multiply the coefficients and add the exponents:

`V = 1 xx 10^10 m^2 * 5.08 xx 10^(-2) m = 5.08 xx 10^8 m^3`

Given that there are 1 x 10 3 liters in a cubic meter we can make the following conversion:

`1 xx 10^3 L * 5.08 xx 10^8 m^3 = 5.08 xx 10^11 L`

Step 5. Check your units and your answer - do they make sense?

`V = 4/3 pi r^3`

Using this equation, plug in the radius (r) of the dust grains.

`V = 4/3 pi (2 xx10^(-6))^3m^3`

Notice the (-6) exponent is cubed. When you take an exponent to an exponent, you need to multiply the two terms

`V = 4/3 pi (8 xx10^(-18)m^3)`

Then, multiple the cubed radius times pi and 4/3

`V = 3.35 xx 10^(-17) m^3`

`m = 3.35 xx 10^(-17) m^3 * 3300 (kg)/m^3`

Notice in the equation above that the m 3 terms cancel each other out and you are left with kg

`m = 1.1 xx 10^(-13) kg`

`V = 4/3 pi (2.325 xx10^(15) m)^3`

`V = 5.26 xx10^(46) m^3`

Number of dust grains = `5.26 xx10^(46) m^3 xx 0.001` grains/m 3

Number of dust grains = `5.26xx10^43 "grains"`

Total mass = `1.1xx10^(-13) (kg)/("grains") * 5.26xx10^43 "grains"`

Notice in the equation above the 'grains' terms cancel each other out and you are left with kg

Total mass = `5.79xx10^30 kg`

If you feel comfortable with this topic, you can go on to the assessment . Or you can go back to the Scientific Notation explanation page .

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Scientific Notation Word Problems

This lesson is about six carefully chosen scientific notation word problems along with their solutions.

Example #1:

Jupiter is the largest planet in our solar system and its mass in scientific notation is about 1.9 × 10 27 . Write the mass in standard notation.

1.9 × 10 27 = 1.9 × 10 × 10 26 = 19 × 10 26 = 1900000000000000000000000000

The mass of Jupiter in standard notation is 1900000000000000000000000000

Example #2:

It is predicted that the world population in 2025 may reach eight billion. Write eight billion in scientific notation.

Eight billion = 8000000000 = 8 × 10 9

Example #3:

The oxygen atom has a mass of about 0.00000000000000000000000003 kg. Write the mass of the oxygen atom in scientific notation.

0.00000000000000000000000003 = 3 × 10 -26 kg

Example #4:

The following masses which are parts of an atom are measured in grams. Order the parts of an atom from greatest to least mass.

electron: 9.1096 × 10 -28 , neutron: 1.6749 × 10 -24 , proton: 1.6726 × 10 -24

The smallest mass is since 9.1096 × 10 -28 since -28 will create more zeros after the decimal point.

1.6749 × 10 -24 and 1.6726 × 10 -24 have the same exponent. So we just need to compare 1.6749 to 1.6726.

Since 1.6726 is smaller than 1.6749, 1.6726 × 10 -24 is smaller than 1.6749 × 10 -24

From greatest to least mass, we get: 1.6749 × 10 -24 > 1.6726 × 10 -24 > 9.1096 × 10 -28

Example #5:

The weight of an adult male elephant is about 5600 kg. What is the weight in scientific notation?

Solution: In scientific notation the weight of an adult male elephant is 5.6 × 10 3

Example #6:

A computer can perform about 5 × 10 8 instructions per second. How many instructions is that per hour? Express the answer in scientific notation.

1 hour is equal to 3600 seconds. Therefore, we need to multiply 5 × 10 8 by 3600

3600 × 5 × 10 8 = 36 × 10 2 × 5 × 10 8

3600 × 5 × 10 8 = 36 × 5 × 10 2 × 10 8

3600 × 5 × 10 8 = 180 × 10 10

3600 × 5 × 10 8 = 1.80 × 10 12

3600 × 5 × 10 8 = 1.8 × 10 12

The computer can perform about 1.8 × 10 12 instructions per hour

Scientific notation

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HiSET: Math : Solve problems using scientific notation

Study concepts, example questions & explanations for hiset: math, all hiset: math resources, example questions, example question #1 : solve problems using scientific notation.

Simplify the following expression using scientific notation.