Question

How do you simplify ${x^2}{y^{ - 4}} \times {x^3}{y^2}$ and write it using only positive exponents?

Answer 1

Hint: In this question, we want to multiply the expression with the exponent. For that, we will combine the like terms. Then we will apply the product of power property which states that if we multiply powers with the same base we just have to add the exponents. The negative exponents are the reciprocals of the positive exponents.
Therefore, the formulas used are:
${x^a} \times {x^b} = {x^{a + b}}$
${x^{ - a}} = \dfrac{1}{{{x^a}}}$

Complete step by step solution:
In this question, we want to multiply the given expression.
$ \Rightarrow {x^2}{y^{ - 4}} \times {x^3}{y^2}$
First, we will combine the like terms together.
That is equal to,
$ \Rightarrow \left( {{x^2} \times {x^3}} \right)\left( {{y^{ - 4}} \times {y^2}} \right)$...(1)
Now, let us apply multiplication on ${x^2}$ and ${x^3}$.
According to the product of power property, when we multiply powers with the same base we just have to add the exponents.
Therefore, the formula is: ${x^a} \times {x^b} = {x^{a + b}}$
In this question, we want to multiply ${x^2}$ and ${x^3}$.
Substitute the values in the product of the power property formula.
$ \Rightarrow {x^2} \times {x^3} = {x^{2 + 3}}$
The addition of 2 and 3 is 5.
 $ \Rightarrow {x^2} \times {x^3} = {x^5}$
 Now, let us apply multiplication on ${y^{ - 4}}$ and ${y^2}$.
According to the product of power property, when we multiply powers with the same base we just have to add the exponents.
Therefore, the formula is: ${x^a} \times {x^b} = {x^{a + b}}$
In this question, we want to multiply ${y^{ - 4}}$ and ${y^2}$.
Substitute the values in the product of the power property formula.
$ \Rightarrow {y^{ - 4}} \times {y^2} = {y^{ - 4 + 2}}$
The addition of -4 and 2 is -2.
 $ \Rightarrow {y^{ - 4}} \times {y^2} = {y^{ - 2}}$
Negative exponents are the reciprocals of the positive exponents.
${x^{ - a}} = \dfrac{1}{{{x^a}}}$
So,
$ \Rightarrow {y^{ - 4}} \times {y^2} = \dfrac{1}{{{y^2}}}$
Put these values in equation (1).
$ \Rightarrow \left( {{x^2} \times {x^3}} \right)\left( {{y^{ - 4}} \times {y^2}} \right) = \dfrac{{{x^5}}}{{{y^2}}}$

Hence, the solution of ${x^2}{y^{ - 4}} \times {x^3}{y^2}$ is $\dfrac{{{x^5}}}{{{y^2}}}$.

Note:
Some exponent properties are as below.
Product of power property: ${x^a} \times {x^b} = {x^{a + b}}$
Power to a power property: ${\left( {{x^a}} \right)^b} = {x^{ab}}$
Power of a product property: ${\left( {xy} \right)^a} = {x^a}{y^a}$
The quotient of power property: $\dfrac{{{x^a}}}{{{x^b}}} = {x^{a - b}}$
Power of a quotient property: ${\left( {\dfrac{x}{y}} \right)^a} = \dfrac{{{x^a}}}{{{y^a}}}$
Negative exponents are the reciprocals of the positive exponents.
Therefore, ${x^{ - a}} = \dfrac{1}{{{x^a}}}$ and
 ${x^a} = \dfrac{1}{{{x^{ - a}}}}$