Question

Use the law of exponents to rewrite and simplify the expression.

$\frac{x^{2\text{n}}·x^{3\text{n}-1}}{x^{\text{n}+2}}$

Answer 1

To simplify the given expression:

Using the law of exponents to simplify the given expression in below:

$\begin{array}{rcl}\frac{x^{2\text{n}}·x^{3\text{n}-1}}{x^{\text{n}+2}}& =& \frac{x^{2\text{n}+3\text{n}-1}}{x^{\text{n}+2}}\left[\text{Product rule}:{\text{a}}^{\text{m}}\times {\text{a}}^{\text{n}}={\text{a}}^{\text{m+n}}\right]\\ & \Rightarrow & \frac{x^{2\text{n}}·x^{3\text{n}-1}}{x^{\text{n}+2}}=\frac{x^{5\text{n}-1}}{x^{\text{n}+2}}\left[\text{Simplifying}\right]\\ & \Rightarrow & \frac{x^{2\text{n}}·x^{3\text{n}-1}}{x^{\text{n}+2}}=x^{5\text{n}-1-\left(\text{n}+2\right)}\left[\text{Quotient rule}:{\text{a}}^{\text{m}}\times {\text{a}}^{\text{n}}={\text{a}}^{\text{m+n}}\right]\\ \therefore \frac{x^{2\text{n}}·x^{3\text{n}-1}}{x^{\text{n}+2}}& =& x^{4\text{n}-3}\left[\text{Simplifying}\right]\end{array}$

Hence, the answer is ${\mathit{x}}^{\mathbf{4}\mathbf{\text{n}}\mathbf{-}\mathbf{3}}$.