Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 5 - Number Theory and the Real Number System - 5.6 Exponents and Scientific Notation - Exercise Set 5.6 - Page 320: 109

Answer

\[8\times {{10}^{-5}}\].

Work Step by Step

If a number is expressed as \[a\times {{10}^{n}}\], where a is a number greater than or equal to 1 and less than 10, and n is an integer, then the number is said to be expressed in scientific notation. Hence all the four numbers, \[1.6\times {{10}^{4}}\], \[7.2\times {{10}^{-3}}\],\[3.6\times {{10}^{8}}\] and \[4\times {{10}^{-3}}\], involved in the given expression are expressed in scientific notation. Now, to compute the given expression, break the computation into three parts. First, compute the multiplications\[\left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)\] and \[\left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)\] separately, and then divide the result of the first multiplication\[\left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)\] by the result of the second multiplication \[\left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)\]. Then, perform the multiplication \[\left( a\times b \right)\] as the usual multiplication of any two numbers is performed and the multiplication \[\left( {{10}^{x}}\times {{10}^{y}} \right)\] using the product rule for exponents, explained below. And then write the two results obtained after performing the above two multiplications, together, with the symbol \[\times \] between them. Now, apply the procedure mentioned above and the product rule for exponents to compute the required multiplications: \[\begin{align} & \left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)=\left( 1.6\times 7.2 \right)\left( {{10}^{4}}\times {{10}^{-3}} \right) \\ & =\left( 11.52 \right)\left( {{10}^{4-3}} \right) \\ & =11.52\times {{10}^{1}} \end{align}\] Hence, \[\left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)=11.52\times {{10}^{1}}\]. Similarly, \[\begin{align} & \left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)=\left( 3.6\times 4 \right)\left( {{10}^{8}}\times {{10}^{-3}} \right) \\ & =\left( 14.4 \right)\left( {{10}^{8-3}} \right) \\ & =14.4\times {{10}^{5}} \end{align}\] Hence, \[\left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)=14.4\times {{10}^{5}}\]. Now, divide the result of the first multiplication\[\left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)\] by the result of the second multiplication\[\left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)\]: \[\frac{\left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)}{\left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)}=\frac{11.52\times {{10}^{1}}}{14.4\times {{10}^{5}}}\] Then, perform the division \[\left( \frac{a}{b} \right)\] as the usual division of any two numbers is performed and the division \[\left( \frac{{{10}^{x}}}{{{10}^{y}}} \right)\] using the quotient rule for exponents, explained below. And then write the two results obtained after performing the above two divisions, together, with the symbol \[\times \] between them. According to the quotient rule for exponents: \[\frac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}\] Now, apply the procedure mentioned above and the quotient rule for exponents to compute the required division: \[\begin{align} & \frac{11.52\times {{10}^{1}}}{14.4\times {{10}^{5}}}=\left( \frac{11.52}{14.4} \right)\times \left( \frac{{{10}^{1}}}{{{10}^{5}}} \right) \\ & =\left( \frac{\left( \right)\left( 0.8 \right)}{} \right)\times \left( {{10}^{1-5}} \right) \\ & =0.8\times {{10}^{-4}} \end{align}\] The result obtained is still not expressed in scientific notation because \[0.8\] is less than 1. Note that \[0.8\] can be rewritten as \[8\times {{10}^{-1}}\], where \[{{10}^{-1}}=\frac{1}{10}\]. Hence, \[\begin{align} & 0.8\times {{10}^{-4}}=\left( 8\times {{10}^{-1}} \right)\times {{10}^{-4}} \\ & =8\times \left( {{10}^{-1}}\times {{10}^{-4}} \right) \\ & =8\times \left( {{10}^{-1-4}} \right) \\ & =8\times {{10}^{-5}} \end{align}\] Hence, \[\begin{align} & \frac{\left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)}{\left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)}=\frac{11.52\times {{10}^{1}}}{14.4\times {{10}^{5}}} \\ & =0.8\times {{10}^{-4}} \\ & =8\times {{10}^{-5}} \end{align}\]
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