## Thinking Mathematically (6th Edition)

$8\times {{10}^{-5}}$.
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}