Scientific notation is a way of writing very large or small numbers in a way that does not use as much space.
The way numbers are written in scientific notation is as follows:
\(a \times 10^b\)
Where \(a\) is a number that is greater than or equal to \(1\) but less than \(10\), while \(b\) is an integer.
If the number is particularly large, then \(b\) will be greater than \(\boldsymbol{0}\); if the number is particularly small, then \(b\) will be less than \(\boldsymbol{0}\).
You may be wondering, “Why use scientific notation at all?” We use scientific notation to make an incredibly large or incredibly small number easier to read.
For example, the speed of light is close to \(300,000,000\) metres per second, but you will find many people writing it as \(3108\) metres per second as it requires less work to write and is much easier to understand than having a ton of zeroes.
Speaking of measurements, notice the prefixes for units of measurement such as metres or grams. Until this point, you will have seen measurements such as CENTImetres, KILOgrams, or KILOmetres. You can actually translate these prefixes into scientific notation so that you can remove the prefix altogether.
For example, one kilogram can be written as \(1 \times 10^3\) grams.
The chart below will show you measurement prefixes, their symbol, and their counterparts in scientific notation. (It will not cover all of them, but just the ones you may frequently see):
| Prefix | Symbol | Numerical Form | Exponential Form |
|---|---|---|---|
| Tera- | T | \(1,000,000,0000,0000\) | \(10^{12}\) |
| Giga- | G | \(1,000,000,0000\) | \(10^9\) |
| Mega- | M | \(1,000,000\) | \(10^6\) |
| Kilo- | k | \(1,000\) | \(10^3\) |
| Hecto- | h | \(100\) | \(10^2\) |
| Deca- | da | \(10\) | \(10^1\) |
| \(1\) | \(10^0\) | ||
| Deci- | d | \(0.1\) | \(10^{-1}\) |
| Centi- | c | \(0.01\) | \(10^{-2}\) |
| Milli- | m | \(0.001\) | \(10^{-3}\) |
| Micro- | \(\mu\) | \(0.000001\) | \(10^{-6}\) |
| Nano- | n | \(0.000000001\) | \(10^{-9}\) |
Write \(643,000,000\) in scientific notation.
First, we must move the decimal so that it is between the first two digits (in this case, between \(6\) and \(4\)) resulting in the following number:
\(6.43000000\)
Next, we can remove all of those zeroes as they are now unnecessary. We now have a decimal after all. This results in the following number:
\(6.43\)
Now we can add the aspect of \(10^b\). The number represented by \(b\) will be how many digits we shifted the decimal to the left. In this case, \(b = 8\). And as we are converting a large number, \(b\) is going to be positive. Thus, we get the value:
\(6.43 \times 10^8\)
Therefore, we can determine that \(643,000,000\) written in scientific notation is \(\boldsymbol{6.43 \times 10^8}\).
Write \(0.00000963\) in scientific notation.
Since we are working with a small number instead of a large number, we can move the decimal so that it is just after the first non-zero digit. In this case, it is \(9\), resulting in the following number:
\(9.63\)
Similar to the previous example, we add \(10^b\) where \(b\) will represent how many digits we had to shift the decimal. With small numbers like this, because we moved the decimal over the other way, \(b\) is going to be negative. Since we moved the decimal over \(6\) digits, then \(b = -6\). Thus, we get the following value:
\(9.63 \times 10^{-6}\)
Therefore, we can determine that \(0.00000963\) written in scientific notation is \(\boldsymbol{9.63 \times 10^{-6}}\).
\(7,240,000\)
Since the number is greater than \(1\), we must move the decimal to the right so that it is between the first two non-zero digits (in this case, between \(7\) and \(2\)). This results in the following number:
\(7.24\)
Next, we can determine the value of \(b\). Since we shifted the decimal point to the left, we can determine that \(b\) will be negative. Additionally, since we shifted \(b\) over \(6\) decimal places, we can determine \(b = 6\).
Thus, we get the following value:
\(7.24 \times 10^6\)
Therefore, we can determine that \(7,240,000\) written in scientific notation is \(\boldsymbol{7.24 \times 10^6}\).
\(0.0000964\)
First, we can calculate the value of \(a\). Since the number is less than \(1\), we must move the decimal to the right so that it is between the first two non-zero digits (in this case, between \(9\) and \(6\)). Therefore, we can determine \(a = 9.64\)
Next, we can calculate the value of \(b\). Since we shifted the decimal point to the left, we can determine that \(b\) will be negative. Additionally, since we shifted \(b\) over \(5\) decimal places, we can determine \(b = -5\).
Thus, we get the following value:
\(9.64 \times 10^{-5}\)
Therefore, we can determine that \(7,240,000\) written in scientific notation is \(\boldsymbol{9.64 \times 10^{-5}}\).
Write \(8.93 \times 10^6\) in standard form.
Converting a number like this is quite simple. You simply need to move the decimal to the right and fill in the empty spaces with zeros. The number of times you move the decimal is the value that is the power to \(10\) which is \(6\). Because the number is positive, this means we move the decimal point to the left over \(6\) digits.
This results in the following value:
\(8,930,000\)
Therefore, we can determine that \(8.93 \times 10^6\) written in standard form is \(\boldsymbol{8,930,000}\).
Write the number \(5.27 \times 10-{3}\) in standard form.
Much like the previous example, you need to move the decimal based on the power of \(10\) which is \(-3\). Since this number is negative, this means we move the decimal point 3 decimal places to the right.
This results in the following value:
\(0.00527\)
Therefore, we can determine that \(5.27 \times 10-{3}\) written in standard form is \(\boldsymbol{0.00527}\).
\(8.12 \times 10^{-7}\)
Since \(b\) is negative, we know that this number is less than \(1\). Additionally, since \(b\) is \(-7\), this means we will move the decimal \(7\) decimal places to the left, filling the empty spaces with \(0\).
This results in the following value:
\(0.000000812\)
Therefore, we can determine that \(8.12 \times 10{-7}\) written in standard form is \(\boldsymbol{0.000000812}\).
\(5 \times 10^6\)
Since \(b\) is positive, we know that this number is greater than \(1\). Additionally, since \(b\) is \(6\), this means we will move the decimal \(6\) decimal places to the right, filling the empty spaces with \(0\).
This results in the following value:
\(5,000,000\)
Therefore, we can determine that \(5 \times 10^6\) written in standard form is \(\boldsymbol{5,000,000}\).
In order to determine how many journeys to the Moon would be equal to journeys to the Sun, we must divide the average distance between the Earth and the Sun by the average distance between the Earth and Moon. We can represent this as such:
\(\cfrac{1.50 \times 10^{11}\;[\text{m}]}{3.84 \times 10^8\;[\text{m}]}\)
Although we can plug these numbers into a calculator, we can first simplify the expression by reducing the exponents and cancelling out units:
\(= \cfrac{1.50 \times 10^{11-8}\;[\cancel{\text{m}}]}{3.84\;[\cancel{\text{m}}]}\)
\(= \cfrac{1.50 \times 10^3}{3.84}\)
Since we cannot reduce this expression any further, we can plug it into a calculator to determine the final result:
\(= \cfrac{1500}{3.84}\)
\(= 390.625\)
Finally, we can rewrite this number in scientific notation. In order to do so, we can shift the decimal \(2\) decimal places to the left. Since the number is greater than \(1\) then we know that \(b\) is positive.
Thus, after rounding the \(a\)-value to the nearest hundredth, we get the following value:
\(3.91 \times 10^2\)
Therefore, we can determine the journey from the Earth to the Sun would be equivalent to about \(\boldsymbol{3.91 \times 10^2}\) journeys to the moon.