Simplify the expression: $\left( {5x{{10}^2}} \right)\left( {3x{{10}^{ - 3}}} \right)$
Answer
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Hint: Simplifying the expression includes the grouping of the similar terms. Simplify makes an expression easy to understand and easy to solve. In this problem, we have to simplify the given expression i.e. $\left( {5x{{10}^2}} \right)\left( {3x{{10}^{ - 3}}} \right)$. To simplify we need to group the numbers, the variables and the exponential numbers together.
Complete step-by-step solution:
To simplify$\left( {5x{{10}^2}} \right)\left( {3x{{10}^{ - 3}}} \right)$ ,
Step (a). We will open the brackets by placing a multiplication sign between it and now it will look like,
$ \Rightarrow 5x{10^2} \times 3x{10^{ - 3}}$
Step (b). If there is no sign placed between the number, the variable and the exponential number, then by default, there is a multiplication sign, now we will place the multiplication sign between them.
$ \Rightarrow 5 \times x \times {10^2} \times 3 \times x \times {10^{ - 3}}$
Step (c). Now, we will place the numbers together.
$ \Rightarrow 5 \times 3 \times x \times {10^2} \times x \times {10^{ - 3}}$
Step (d). On multiplying the numbers the expression become,
$ \Rightarrow 15 \times x \times {10^2} \times x \times {10^{ - 3}}$
Step (e). Now, we will place the variables and the exponents together, we get,
$ \Rightarrow 15 \times x \times x \times {10^2} \times {10^{ - 3}}$
Step (f). If there is no exponent (power) on the variable then the power is assumed as 1, and if the two similar terms are multiplied then their power is added to give the answer. Now, the expression become,
$ \Rightarrow 15 \times {x^{1 + 1}} \times {10^{2 - 3}}$
Step (g). On further solving, we get,
$ \Rightarrow 15 \times {x^2} \times {10^{ - 1}}$
Step (h). If the term has a negative power, then it is written inversely. Then the expression will become,
$ \Rightarrow 15 \times {x^2} \times \dfrac{1}{{10}}$
Hence, the simplified expression is $1.5{x^2}$.
Note: The properties of the exponents that are used to simplify the given expression is
(i)${x^a} \times {x^b} = {x^{a + b}}$
For example, ${2^2} \times {2^5} = {2^7}$
(ii)${x^{ - y}} = \dfrac{1}{{{x^y}}}$
For example, ${5^{ - 1}} = \dfrac{1}{5}$
These properties help us to rewrite the exponents in different ways.
Complete step-by-step solution:
To simplify$\left( {5x{{10}^2}} \right)\left( {3x{{10}^{ - 3}}} \right)$ ,
Step (a). We will open the brackets by placing a multiplication sign between it and now it will look like,
$ \Rightarrow 5x{10^2} \times 3x{10^{ - 3}}$
Step (b). If there is no sign placed between the number, the variable and the exponential number, then by default, there is a multiplication sign, now we will place the multiplication sign between them.
$ \Rightarrow 5 \times x \times {10^2} \times 3 \times x \times {10^{ - 3}}$
Step (c). Now, we will place the numbers together.
$ \Rightarrow 5 \times 3 \times x \times {10^2} \times x \times {10^{ - 3}}$
Step (d). On multiplying the numbers the expression become,
$ \Rightarrow 15 \times x \times {10^2} \times x \times {10^{ - 3}}$
Step (e). Now, we will place the variables and the exponents together, we get,
$ \Rightarrow 15 \times x \times x \times {10^2} \times {10^{ - 3}}$
Step (f). If there is no exponent (power) on the variable then the power is assumed as 1, and if the two similar terms are multiplied then their power is added to give the answer. Now, the expression become,
$ \Rightarrow 15 \times {x^{1 + 1}} \times {10^{2 - 3}}$
Step (g). On further solving, we get,
$ \Rightarrow 15 \times {x^2} \times {10^{ - 1}}$
Step (h). If the term has a negative power, then it is written inversely. Then the expression will become,
$ \Rightarrow 15 \times {x^2} \times \dfrac{1}{{10}}$
Hence, the simplified expression is $1.5{x^2}$.
Note: The properties of the exponents that are used to simplify the given expression is
(i)${x^a} \times {x^b} = {x^{a + b}}$
For example, ${2^2} \times {2^5} = {2^7}$
(ii)${x^{ - y}} = \dfrac{1}{{{x^y}}}$
For example, ${5^{ - 1}} = \dfrac{1}{5}$
These properties help us to rewrite the exponents in different ways.
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