4.1: Scientific Notation
- Page ID
- 20038
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Scientific notation is used to make very large numbers such as 4,895,000,000 or very small numbers such as 0.0000073 easier to use. In the Water industry, we use very large numbers when referring to the volume of water in large tanks or very small numbers when referring to pollutants per gallon of water.
Scientific Notation Form:
\[\underbrace{(a\ decimal)}_{between\ 0\ and\ 10\ \left(not\ including\ 10\right)} \times {10}^{power}\]
To convert from scientific notation to the actual number
- For a positive exponent on the 10, move the decimal place to the right the equivalent number of spaces as the power
- For a negative exponent on the 10, move the decimal place to the left the equivalent number of spaces as the absolute value of the power
Convert the scientific notation to the actual number: \(8.735 \times {10}^7\)
Solution
Since the power of 10 is positive 7, we will move the decimal place 7 units to the RIGHT to obtain: 87,350,000
Convert the scientific notation to the actual number: \(2.356 \times {10}^{-4}\)
Solution
Since the power of 10 is negative 4, we will move the decimal place 4 units to the LEFT to obtain: 0.2356
To convert a positive number to scientific notation
- Move the decimal place to the right of the first non-zero digit. This will be the decimal number portion of the scientific notation.
- If the decimal place was moved to the left, use a positive power of 10 based on the number of places the decimal was moved
- If the decimal place was moved to the right, use a negative power of 10 based on the number of places the decimal was moved
Convert 567,900,000 to scientific notation.
Solution
Currently, the decimal is understood to be after the last digit since this is a whole number, so move the decimal to the left 8 places so the decimal is between 5 and 6, hence the scientific notation is \(5.679 \times {10}^8\)
Convert 0.00032 to scientific notation.
Solution
Move the decimal to the right 4 places so the decimal is between 3 and 2 to create a decimal number of 3.2 and the power of 10 would then be negative 4 since we moved the decimal to the right 4 places. The scientific notation would be \(3.2 \times {10}^{-4}\)
Numbers that are larger than 1 will have a positive power of 10 in scientific notation and numbers that are less than 1 (but still positive) will have a negative power of 10 in scientific notation.
Display of scientific notation on scientific calculators
To multiply and divide numbers in scientific notation, we can multiply (or divide) the decimals and multiply (or divide) the powers of 10, then simplify by rewriting into the proper scientific notation form. To do this, we will need to understand some rules of exponents.
Rule of Exponents
1. When multiplying expressions with the same base, we keep the base and add the exponents:
\[a^x\cdot a^y=a^{x+y}\]
2. When dividing expressions with the same base, we keep the base and subtract the exponents:
\[\frac{a^x}{a^y}=a^{x-y}\]
3. Negative exponents become positive exponents if we move the expression to the opposite side of a fraction:
\[a^{-x}=\frac{1}{a^x} \,\,\, \text{and} \,\,\, \frac{1}{a^{-x}}=a^x\]
4. A non-zero expression raised to an exponent of zero is equivalent to 1:
\[a^0=1\]
Multiply and simplify using scientific notation: \((2.45\times {10}^6)(3.23\times {10}^{-15})\)
Solution
First, we can rearrange the multiplication to obtain:
\[\left(2.45\times {10}^6\right)\left(3.23\times {10}^{-15}\right)\]
\[=\left(2.45\right)\left(3.23\right)\times ({10}^6)({10}^{-15})\]
Next, we can multiply the decimals, then multiply the powers of 10 using an exponent rule:
\[=7.9135\times {10}^{6+\left(-15\right)}\]
\[=7.9135\times {10}^{-9}\]
Since the decimal value of 7.9135 is between 0 and 10 (not including 10), then this is the proper scientific notation form.
Multiply and simplify using scientific notation: \((8.7\times {10}^{-6})(2.5\times {10}^{12})\)
Solution
First, we can rearrange the multiplication to obtain:
\[(8.7\times {10}^{-6})(2.5\times {10}^{12})=\left(8.7\right)\left(2.5\right)\times ({10}^{-6})({10}^{12})\]
Next, we can multiply the decimals, then multiply the powers of 10 using an exponent rule:
\[=21.75\times {10}^{-6+12}\]
\[=21.75\times {10}^6\]
Since the decimal value of 21.75 is not between 0 and 10 (not including 10), we need to convert it to proper scientific notation form and simplify further
\[21.75\times {10}^6=\left(2.175\times {10}^1\right)\times {10}^6\]
\[\ \ \ \ \ \ =2.175\times {10}^{1+6}\]
\[=2.175\times {10}^7\]
Divide and simplify using scientific notation: \(\frac{4.125\times {10}^{13}}{7.5\times {10}^{-2}}\)
Solution
First, we divide the decimals, then divide the powers of 10 using an exponent rule:
\[\frac{4.125\times 10^{13}}{7.5\times 10^{-2}}=\frac{4.125}{7.5} \times \frac{10^{13}}{10^{-2}}\]
\[=0.55\times {10}^{13-\left(-2\right)}\]
\[=0.55\times {10}^{15}\]
Since the decimal value of 0.55 is not between 0 and 10 (not including 10), we need to convert it to proper scientific notation form and simplify further
\[0.55\times {10}^{15}=\left(5.5\times {10}^{-1}\right)\times {10}^{15}\]
\[=5.5\times {10}^{-1+15}\]
\[=5.5\times {10}^{14}\]