## Thinking Mathematically (6th Edition)

$2.5\times {{10}^{-3}}$.
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 three numbers, $5\times {{10}^{3}}$, $1.2\times {{10}^{-4}}$ and $2.4\times {{10}^{2}}$, involved in the given expression are expressed in scientific notation. Now, to compute the given expression, break the computation into two parts. First, compute the multiplication$\left( 5\times {{10}^{3}} \right)\left( 1.2\times {{10}^{-4}} \right)$, and then divide the result of the multiplication by$\left( 2.4\times {{10}^{2}} \right)$. To perform the multiplication of two numbers which are expressed in scientific notation, first write the multiplication as separate multiplications of numbers involving exponents of 10 and that of the other parts of the given numbers as shown: $\left( a\times {{10}^{x}} \right)\left( b\times {{10}^{y}} \right)=\left( a\times b \right)\left( {{10}^{x}}\times {{10}^{y}} \right)$ 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 multiplication: \begin{align} & \left( 5\times {{10}^{3}} \right)\left( 1.2\times {{10}^{-4}} \right)=\left( 5\times 1.2 \right)\left( {{10}^{3}}\times {{10}^{-4}} \right) \\ & =\left( 6.0 \right)\left( {{10}^{3-4}} \right) \\ & =6\times {{10}^{-1}} \end{align} Hence, $\left( 5\times {{10}^{3}} \right)\left( 1.2\times {{10}^{-4}} \right)=6\times {{10}^{-1}}$. Now, divide the result obtained in the above multiplication by $\left( 2.4\times {{10}^{2}} \right)$: Hence, \begin{align} & \left( 5\times {{10}^{3}} \right)\left( 1.2\times {{10}^{-4}} \right)\div \left( 2.4\times {{10}^{2}} \right)=\left( 6\times {{10}^{-1}} \right)\div \left( 2.4\times {{10}^{2}} \right) \\ & =\frac{6\times {{10}^{-1}}}{2.4\times {{10}^{2}}} \end{align} 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. Now, apply the procedure mentioned above and the quotient rule for exponents to compute the required division: \begin{align} & \frac{6\times {{10}^{-1}}}{2.4\times {{10}^{2}}}=\left( \frac{6}{2.4} \right)\times \left( \frac{{{10}^{-1}}}{{{10}^{2}}} \right) \\ & =\left( \frac{\left( \right)\left( 2.5 \right)}{} \right)\times \left( {{10}^{-1-2}} \right) \\ & =2.5\times {{10}^{-3}} \end{align} Hence, \begin{align} & \left( 5\times {{10}^{3}} \right)\left( 1.2\times {{10}^{-4}} \right)\div \left( 2.4\times {{10}^{2}} \right)=\left( 6\times {{10}^{-1}} \right)\div \left( 2.4\times {{10}^{2}} \right) \\ & =\frac{6\times {{10}^{-1}}}{2.4\times {{10}^{2}}} \\ & =2.5\times {{10}^{-3}} \end{align}