Question

How do you simplify $\left( {2{x^2}y} \right)\left( {5{x^{ - 2}}{y^3}} \right)$ and write it only using positive exponents?

Answer 1

Hint: We will first multiply both the terms in multiplication and remove the parentheses. Then, we will use the fact that ${u^a}{u^b} = {u^{a + b}}$ on both x and y and thus we have the required answer.

Complete step by step solution:
We are given that we are required to simplify $\left( {2{x^2}y} \right)\left( {5{x^{ - 2}}{y^3}} \right)$ and write it only using positive exponents.
Now, we can write the given expression $\left( {2{x^2}y} \right)\left( {5{x^{ - 2}}{y^3}} \right)$ as written follows:-
$ \Rightarrow \left( {2{x^2}y} \right)\left( {5{x^{ - 2}}{y^3}} \right) = 2{x^2}y \times 5{x^{ - 2}}{y^3}$
Simplifying the right hand side of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow \left( {2{x^2}y} \right)\left( {5{x^{ - 2}}{y^3}} \right) = 10\left( {{x^2}y \times {x^{ - 2}}{y^3}} \right)$
Now, we know that we have a fact given by the following expression with us:-
$ \Rightarrow {u^a}{u^b} = {u^{a + b}}$, where u, a and b are real numbers.
Therefore, we will now obtain the following equation with us:-
$ \Rightarrow \left( {2{x^2}y} \right)\left( {5{x^{ - 2}}{y^3}} \right) = 10{x^{2 - 2}}{y^{3 + 1}}$
Simplifying the powers on the right hand side, we will then obtain the following equation with us:-
$ \Rightarrow \left( {2{x^2}y} \right)\left( {5{x^{ - 2}}{y^3}} \right) = 10{x^0}{y^4}$
Now, we know that, for any real number a which is not equal to 0, we have ${a^0} = 1$.
Therefore, we have the following equation with us now:-

$ \Rightarrow \left( {2{x^2}y} \right)\left( {5{x^{ - 2}}{y^3}} \right) = 10{y^4}$

Thus, we have an expression with only positive exponents.

Note:
1) The students must note that 0 is not a positive number. It is just a non – zero number but not a positive one. Positive number is strictly greater than zero. Therefore, any positive number can never be equal to zero. Therefore, we just remove the x portion with the 0 exponent because that would not have been classified as a positive exponent.
2) The students must also keep in mind that while using the identity that says that ${a^0} = 1$, one must make sure of the fact that $a \ne 0$ and $a \in \mathbb{R}$. Here, in the given question, we knew that x is not equal to 0, because if x would have been equal to 0, then the expression itself would have become zero in the initial only.