Question

How do you simplify \[\dfrac{2{{x}^{-3}}\times {{x}^{5}}}{3{{x}^{2}}}\]?

Answer 1

Hint: we can simplify this expression using basic exponent formulas. First we have to simplify the numerator using the formulas. And then we will simplify both numerator and denominator to get the simplest form of the given expression.

Complete step by step solution:
Before going to solve let us know a few basic formulas of exponents and powers. They are
\[{{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}\]
\[{{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}\]
\[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\]
\[{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}\]
\[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\]
These are the basic formulas of exponents and we will use them in the problem where ever needed
Given equation is
\[\dfrac{2{{x}^{-3}}\times {{x}^{5}}}{3{{x}^{2}}}\]
First we will simplify the numerator by reducing it into a single term.
Now we have to simplify the numerator and make it a single term.
By seeing the we can see that we can apply the formula
\[{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}\] to simplify it further.
By applying this formula the numerator will become as
\[\Rightarrow \dfrac{2{{x}^{-3+5}}}{3{{x}^{2}}}\]
Now by simplifying by adding the powers it we will get
\[\Rightarrow \dfrac{2{{x}^{2}}}{3{{x}^{2}}}\]
Now we can divide both numerator and denominator with \[{{x}^{2}}\].
After dividing them both we will get
\[\Rightarrow \dfrac{\dfrac{2{{x}^{2}}}{{{x}^{2}}}}{\dfrac{3{{x}^{2}}}{{{x}^{2}}}}\]
By simplifying it we will get
\[\Rightarrow \dfrac{2}{3}\]

So the simplified form of the given expression is \[\dfrac{2}{3}\].

Note: we can solve these types of questions very easily. But we should have good knowledge about formulas of power and we should be able to assess which formula has to be used next. If we have used any other formula it will be difficult to get the solution and we have to be careful with calculations with powers. we have to check whether the terms have the same base or different bases and then apply the formula accordingly.