Question

How do you multiply $\left( {2x{{10}^4}} \right)\left( {3x{{10}^5}} \right)$?

Answer 1

Hint: In the given problem, $x$ is variable and other terms are constant. To solve this problem, first we will rewrite the given expression and then we will use the law of exponent which is given by ${a^m} \times {a^n} = {a^{m + n}}$.

Complete step-by-step answer:
In this problem, we have to find multiplication of two terms $2x{10^4}$ and $3x{10^5}$. That is, we have to find $\left( {2x{{10}^4}} \right) \times \left( {3x{{10}^5}} \right)$. Note that here $x$ is variable and other terms are constant. We know that order does not matter in the multiplication. So, we can write
$\left( {2x{{10}^4}} \right) \times \left( {3x{{10}^5}} \right) = \left( {2 \times 3} \right) \times \left( {{x^1} \times {x^1}} \right) \times \left( {{{10}^4} \times {{10}^5}} \right)$
Now in the second and third brackets of above expression, we are going to use the law of exponent which is given by ${a^m} \times {a^n} = {a^{m + n}}$. So, we can write
$
  \left( {2x{{10}^4}} \right) \times \left( {3x{{10}^5}} \right) = 6 \times \left( {{x^{1 + 1}}} \right) \times \left( {{{10}^{4 + 5}}} \right) \\
   \Rightarrow \left( {2x{{10}^4}} \right) \times \left( {3x{{10}^5}} \right) = 6 \times \left( {{x^2}} \right) \times \left( {{{10}^9}} \right) \\
   \Rightarrow \left( {2x{{10}^4}} \right) \times \left( {3x{{10}^5}} \right) = 6{x^2}\left( {{{10}^9}} \right) \\
 $

Hence, the multiplication of two terms $2x{10^4}$ and $3x{10^5}$ is equal to $6{x^2}\left( {{{10}^9}} \right)$.

Note:
In the given problem, we can observe that there are some terms in which the bases are the same and so we can combine those terms by using the law of exponents. If the bases are different then we cannot combine those terms. For example, when we multiply ${x^2}$ with ${y^3}$ then we cannot combine these two terms because bases $x$ and $y$ are different. If there is multiplication then we can add the powers on the same base. If there is division then we can subtract the powers on the same base. We must remember the other law of exponent which is given by $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$.