# 8.4: Scientific Notation

**At Grade**Created by: CK-12

## Learning Objectives

- Write numbers in scientific notation.
- Evaluate expressions in scientific notation.
- Evaluate expressions in scientific notation using a graphing calculator.

## Introduction

Consider the number six hundred and forty three thousand, two hundred and ninety seven. We write it as 643,297 and each digit’s position has a “value” assigned to it. You may have seen a table like this before:

\begin{align*}& \text{hundred-thousands} \quad \text{ten-thousands} \quad \text{thousands} \quad \text{hundreds} \quad \text{tens} \quad \text{units}\\ & \qquad \quad \ 6 \qquad \qquad \qquad \qquad 4 \qquad \qquad \qquad 3 \qquad \qquad \quad 2 \qquad \quad 9 \qquad \ \ 7\end{align*}

We’ve seen that when we write an exponent above a number, it means that we have to multiply a certain number of copies of that number together. We’ve also seen that a zero exponent always gives us 1, and negative exponents give us fractional answers.

Look carefully at the table above. Do you notice that all the column headings are powers of ten? Here they are listed:

\begin{align*}100,000 &= 10^5\\ 10,000 &= 10^4\\ 1,000 &= 10^3\\ 100 &= 10^2\\ 10 &= 10^1\end{align*}

Even the “units” column is really just a power of ten. ** Unit** means 1, and 1 is \begin{align*}10^0\end{align*}.

If we divide 643,297 by 100,000 we get 6.43297; if we multiply 6.43297 by 100,000 we get 643, 297. But we have just seen that 100,000 is the same as \begin{align*}10^5\end{align*}, so if we multiply 6.43297 by \begin{align*}10^5\end{align*} we should also get 643,297. In other words,

\begin{align*}643,297 = 6.43297 \times 10^5\end{align*}

## Writing Numbers in Scientific Notation

In scientific notation, numbers are always written in the form \begin{align*}a \times 10^b\end{align*} , where \begin{align*}b\end{align*} is an integer and \begin{align*}a\end{align*} is between 1 and 10 (that is, it has exactly 1 nonzero digit before the decimal). This notation is especially useful for numbers that are either very small or very large.

Here’s a set of examples:

\begin{align*}1.07 \times 10^4 &= 10,700\\ 1.07 \times 10^3 &= 1,070\\ 1.07 \times 10^2 &= 107\\ 1.07 \times 10^1 &= 10.7\\ 1.07 \times 10^0 &= 1.07\\ 1.07 \times 10^{-1} &= 0.107\\ 1.07 \times 10^{-2} &= 0.0107\\ 1.07 \times 10^{-3} &= 0.00107\\ 1.07 \times 10^{-4} &= 0.000107\end{align*}

Look at the first example and notice where the decimal point is in both expressions.

So the exponent on the ten acts to move the decimal point over to the right. An exponent of 4 moves it 4 places and an exponent of 3 would move it 3 places.

This makes sense because each time you multiply by 10, you move the decimal point one place to the right. 1.07 times 10 is 10.7, times 10 again is 107.0, and so on.

Similarly, if you look at the later examples in the table, you can see that a negative exponent on the 10 means the decimal point moves that many places to the left. This is because multiplying by \begin{align*}10^{-1}\end{align*} is the same as multiplying by \begin{align*}\frac{1}{10}\end{align*}, which is like dividing by 10. So instead of moving the decimal point one place to the right for every multiple of 10, we move it one place to the left for every multiple of \begin{align*}\frac{1}{10}\end{align*}.

That’s how to convert numbers from scientific notation to standard form. When we’re converting numbers *to* scientific notation, however, we have to apply the whole process backwards. First we move the decimal point until it’s immediately after the first nonzero digit; then we count how many places we moved it. If we moved it to the *left,* the exponent on the 10 is positive; if we moved it to the *right,* it’s negative.

So, for example, to write 0.000032 in scientific notation, we’d first move the decimal five places to the right to get 3.2; then, since we moved it right, the exponent on the 10 should be *negative* five, so the number in scientific notation is \begin{align*}3.2 \times 10^{-5}\end{align*}.

You can double-check whether you’ve got the right direction by comparing the number in scientific notation with the number in standard form, and thinking “Does this represent a *big* number or a *small* number?” A positive exponent on the 10 represents a number bigger than 10 and a negative exponent represents a number smaller than 10, and you can easily tell if the number in standard form is bigger or smaller than 10 just by looking at it.

For more practice, try the online tool at http://hotmath.com/util/hm_flash_movie.html?movie=/learning_activities/interactivities/sciNotation.swf. Click the arrow buttons to move the decimal point until the number in the middle is written in proper scientific notation, and see how the exponent changes as you move the decimal point.

**Example 1**

*Write the following numbers in scientific notation.*

a) 63

b) 9,654

c) 653,937,000

d) 0.003

e) 0.000056

f) 0.00005007

**Solution**

a) \begin{align*}63 = 6.3 \times 10 = 6.3 \times 10^1\end{align*}

b) \begin{align*}9,654 = 9.654 \times 1,000 = 9.654 \times 10^3\end{align*}

c) \begin{align*}653,937,000 = 6.53937000 \times 100,000,000 = 6.53937 \times 10^8\end{align*}

d) \begin{align*}0.003 = 3 \times \frac{1}{1000} = 3 \times 10^{-3}\end{align*}

e) \begin{align*}0.000056 = 5.6 \times \frac{1}{100,000} = 5.6 \times 10^{-5}\end{align*}

f) \begin{align*}0.00005007 = 5.007 \times \frac{1}{100,000} = 5.007 \times 10^{-5}\end{align*}

## Evaluating Expressions in Scientific Notation

When we are faced with products and quotients involving scientific notation, we need to remember the rules for exponents that we learned earlier. It’s relatively straightforward to work with scientific notation problems if you remember to combine all the powers of 10 together. The following examples illustrate this.

**Example 2**

*Evaluate the following expressions and write your answer in scientific notation.*

a) \begin{align*}(3.2 \times 10^6) \cdot (8.7 \times 10^{11})\end{align*}

b) \begin{align*}(5.2 \times 10^{-4}) \cdot (3.8 \times 10^{-19})\end{align*}

c) \begin{align*}(1.7 \times 10^6) \cdot (2.7 \times 10^{-11})\end{align*}

**Solution**

The key to evaluating expressions involving scientific notation is to group the powers of 10 together and deal with them separately.

a) \begin{align*}(3.2 \times 10^6) (8.7 \times 10^{11}) = \underbrace{3.2 \times 8.7}_{27.84} \times \underbrace{10^6 \times 10^{11}}_{10^{17}} = 27.84 \times 10^{17}\end{align*}. But \begin{align*}27.84 \times 10^{17}\end{align*} isn’t in proper scientific notation, because it has more than one digit before the decimal point. We need to move the decimal point one more place to the left and add 1 to the exponent, which gives us \begin{align*}2.784 \times 10^{18}\end{align*}.

b) \begin{align*}(5.2 \times 10^{-4}) (3.8 \times 10^{-19}) = \underbrace{5.2 \times 3.8}_{19.76} \times \underbrace{10^{-4} \times 10^{-19}}_{10^{-23}} = 19.76 \times 10^{-23} = 1.976 \times 10^{-22}\end{align*}

c) \begin{align*}(1.7 \times 10^6) (2.7 \times 10^{-11}) = \underbrace{1.7 \times 2.7}_{4.59} \times \underbrace{10^6 \times 10^{-11}}_{10^{-5}} = 4.59 \times 10^{-5}\end{align*}

When we use scientific notation in the real world, we often round off our calculations. Since we’re often dealing with very big or very small numbers, it can be easier to round off so that we don’t have to keep track of as many digits—and scientific notation helps us with that by saving us from writing out all the extra zeros. For example, if we round off 4,227, 457,903 to 4,200,000,000, we can then write it in scientific notation as simply \begin{align*}4.2 \times 10^9\end{align*}.

When rounding, we often talk of **significant figures** or **significant digits**. Significant figures include

- all nonzero digits
- all zeros that come
*before*a nonzero digit and*after*either a decimal point or another nonzero digit

For example, the number 4000 has one significant digit; the zeros don’t count because there’s no nonzero digit after them. But the number 4000.5 has five significant digits: the 4, the 5, and all the zeros in between. And the number 0.003 has three significant digits: the 3 and the two zeros that come between the 3 and the decimal point.

**Example 3**

*Evaluate the following expressions. Round to 3 significant figures and write your answer in scientific notation.*

a) \begin{align*}(3.2 \times 10^6) \div (8.7 \times 10^{11})\end{align*}

b) \begin{align*}(5.2 \times 10^{-4}) \div (3.8 \times 10^{-19})\end{align*}

c) \begin{align*}(1.7 \times 10^6) \div (2.7 \times 10^{-11})\end{align*}

**Solution**

It’s easier if we convert to fractions and THEN separate out the powers of 10.

a) \begin{align*}(3.2 \times 10^6) \div (8.7 \times 10^{11}) &= \frac{3.2 \times 10^6}{8.7 \times 10^{11}} && - separate \ out \ the \ powers \ of \ 10:\\ & = \frac{3.2}{8.7} \times \frac{10^6}{10^{11}} && - evaluate \ each \ fraction \ (round \ to \ 3 \ s.f.):\\ & = 0.368 \times 10^{(6 - 11)}\\ & = 0.368 \times 10^{-5} && - remember \ how \ to \ write \ scientific \ notation!\\ & = 3.68 \times 10^{-6}\end{align*}

b) \begin{align*}(5.2 \times 10^{-4}) \div (3.8 \times 10^{-19}) & = \frac{5.2 \times 10^{-4}}{3.8 \times 10^{-19}} && - separate \ the \ powers \ of \ 10:\\ & = \frac{5.2}{3.8} \times \frac{10^{-4}}{10^{-19}} && - evaluate \ each \ fraction \ (round \ to \ 3 \ s.f.)\\ & = 1.37 \times 10^{((-4) - (-19))}\\ & = 1.37 \times 10^{15}\end{align*}

c) \begin{align*}(1.7 \times 10^6) \div (2.7 \times 10^{-11}) & = \frac{1.7 \times 10^6}{2.7 \times 10^{-11}} && - next \ we \ separate \ the \ powers \ of \ 10:\\ & = \frac{1.7}{2.7} \times \frac{10^6}{10^{-11}} && - evaluate \ each \ fraction \ (round \ to \ 3 \ s.f.)\\ & = 0.630 \times 10^{(6-(-11))}\\ & = 0.630 \times 10^{17}\\ & = 6.30 \times 10^{16}\end{align*}

Note that we have to leave in the final zero to indicate that the result has been rounded.

## Evaluate Expressions in Scientific Notation Using a Graphing Calculator

All scientific and graphing calculators can use scientific notation, and it’s very useful to know how.

To insert a number in scientific notation, use the **[EE]** button. This is **[2nd] [,]** on some TI models.

For example, to enter \begin{align*}2.6 \times 10^5\end{align*}, enter 2.6 **[EE]** 5. When you hit **[ENTER]** the calculator displays 2.6E5 if it’s set in **Scientific** mode, or 260000 if it’s set in **Normal** mode.

(To change the mode, press the ‘Mode’ key.)

**Example 4**

*Evaluate \begin{align*}(2.3 \times 10^6) \times (4.9 \times 10^{-10})\end{align*} using a graphing calculator.*

**Solution**

Enter 2.3 **[EE]** \begin{align*}6 \times 4.9\end{align*} **[EE]** - 10 and press **[ENTER]**.

The calculator displays 6.296296296E16 whether it’s in Normal mode or Scientific mode. That’s because the number is so big that even in Normal mode it won’t fit on the screen. The answer displayed instead isn’t the precisely correct answer; it’s rounded off to 10 significant figures.

Since it’s a repeating decimal, though, we can write it more efficiently *and* more precisely as \begin{align*}6. \overline{296} \times 10^{16}\end{align*}.

**Example 5**

*Evaluate* \begin{align*}(4.5 \times 10^{14})^3\end{align*} *using a graphing calculator.*

**Solution**

Enter (4.5 **[EE]** \begin{align*}14)^{\land} 3\end{align*} and press **[ENTER]**.

The calculator displays 9.1125E43. The answer is \begin{align*}9.1125 \times 10^{43}\end{align*}.

## Solve Real-World Problems Using Scientific Notation

**Example 6**

*The mass of a single lithium atom is approximately one percent of one millionth of one billionth of one billionth of one kilogram. Express this mass in scientific notation.*

**Solution**

We know that a *percent* is \begin{align*}\frac{1}{100}\end{align*}, and so our calculation for the mass (in kg) is:

\begin{align*}\frac{1}{100} \times \frac{1}{1,000,000} \times \frac{1}{1,000,000,000} \times \frac{1}{1,000,000,000} = 10^{-2} \times 10^{-6} \times 10^{-9} \times 10^{-9}\end{align*}

Next we use the product of powers rule we learned earlier:

\begin{align*}10^{-2} \times 10^{-6} \times 10^{-9} \times 10^{-9} = 10^{((-2) + (-6) + (-9) + (-9))} = 10^{-26} \ kg.\end{align*}

The mass of one lithium atom is approximately \begin{align*}1 \times 10^{-26} \ kg\end{align*}.

**Example 7**

*You could fit about 3 million \begin{align*}E\end{align*}. coli bacteria on the head of a pin. If the size of the pin head in question is* \begin{align*}1.2 \times 10^{-5} \ m^{2}\end{align*}, *calculate the area taken up by one \begin{align*}E\end{align*}. coli bacterium. Express your answer in scientific notation*

**Solution**

Since we need our answer in scientific notation, it makes sense to convert 3 million to that format first:

\begin{align*}3,000,000 = 3 \times 10^6\end{align*}

Next we need an expression involving our unknown, the area taken up by one bacterium. Call this \begin{align*}A\end{align*}.

\begin{align*}& 3 \times 10^6 \cdot A = 1.2 \times 10^{-5} && - since \ 3 \ million \ of \ them \ make \ up \ the \ area \ of \ the \ pin-head\end{align*}

Isolate \begin{align*}A\end{align*}:

\begin{align*}& A = \frac{1}{3 \times 10^6} \cdot 1.2 \times 10^{-5} && - rearranging \ the \ terms \ gives:\\ & A = \frac{1.2}{3} \cdot \frac{1}{10^6} \times 10^{-5} && - then \ using \ the \ definition \ of \ a \ negative \ exponent:\\ & A = \frac{1.2}{3} \cdot 10^{-6} \times 10^{-5} && - evaluate \And \ combine \ exponents \ using \ the \ product \ rule:\\ & A = 0.4 \times 10^{-11} && - but \ we \ can’t \ leave \ our \ answer \ like \ this, \ so \ldots\end{align*}

The area of one bacterium is \begin{align*}4.0 \times 10^{-12} \ m^{2}\end{align*}.

(Notice that we had to move the decimal point over one place to the right, subtracting 1 from the exponent on the 10.)

## Review Questions

Write the numerical value of the following.

- \begin{align*}3.102 \times 10^2\end{align*}
- \begin{align*}7.4 \times 10^4\end{align*}
- \begin{align*}1.75 \times 10^{-3}\end{align*}
- \begin{align*}2.9 \times 10^{-5}\end{align*}
- \begin{align*}9.99 \times 10^{-9}\end{align*}

Write the following numbers in scientific notation.

- 120,000
- 1,765,244
- 12
- 0.00281
- 0.000000027

How many significant digits are in each of the following?

- 38553000
- 2754000.23
- 0.0000222
- 0.0002000079

Round each of the following to two significant digits.

- 3.0132
- 82.9913

Perform the following operations and write your answer in scientific notation.

- \begin{align*}(3.5 \times 10^4) \cdot (2.2 \times 10^7)\end{align*}
- \begin{align*}\frac{2.1 \times 10^9}{3 \times 10^2}\end{align*}
- \begin{align*}(3.1 \times 10^{-3}) \cdot (1.2 \times 10^{-5})\end{align*}
- \begin{align*}\frac{7.4 \times 10^{-5}}{3.7 \times 10^{-2}}\end{align*}
- \begin{align*}12,000,000 \times 400,000\end{align*}
- \begin{align*}3,000,000 \times 0.00000000022\end{align*}
- \begin{align*}\frac{17,000}{680,000,000}\end{align*}
- \begin{align*}\frac{25,000,000}{0.000000005}\end{align*}
- \begin{align*}\frac{0.0000000000042}{0.00014}\end{align*}
- The moon is approximately a sphere with radius \begin{align*}r = 1.08 \times 10^3 \ miles\end{align*}. Use the formula Surface Area \begin{align*}= 4 \pi r^2\end{align*} to determine the surface area of the moon, in square miles. Express your answer in scientific notation, rounded to two significant figures.
- The charge on one electron is approximately \begin{align*}1.60 \times 10^{19}\end{align*} coulombs. One
**Faraday**is equal to the total charge on \begin{align*}6.02 \times 10^{23}\end{align*} electrons. What, in coulombs, is the charge on one Faraday? - Proxima Centauri, the next closest star to our Sun, is approximately \begin{align*}2.5 \times 10^{13}\end{align*} miles away. If light from Proxima Centauri takes \begin{align*}3.7 \times 10^4\end{align*} hours to reach us from there, calculate the speed of light in miles per hour. Express your answer in scientific notation, rounded to 2 significant figures.

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Please Sign In to create your own Highlights / Notes | |||

Show More |