## College Algebra (11th Edition)

$y^{1/2}$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $\dfrac{y^{5/3}\cdot y^{-2}}{y^{-5/6}} .$ $\bf{\text{Solution Details:}}$ Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{y^{\frac{5}{3}+(-2)}}{y^{-5/6}} \\\\= \dfrac{y^{\frac{5}{3}-2}}{y^{-5/6}} .\end{array} To simplify the expression $\dfrac{5}{3}-2 ,$ change the expressions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $3$ and $1$ is $3$ since it is the lowest number that can be exactly divided by the denominators. Multiplying the terms by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to \begin{array}{l}\require{cancel} \dfrac{5}{3}-2\cdot\dfrac{3}{3} \\\\= \dfrac{5}{3}-\dfrac{6}{3} \\\\= \dfrac{5-6}{3} \\\\= \dfrac{-1}{3} \\\\= -\dfrac{1}{3} .\end{array} The expression, $\dfrac{y^{\frac{5}{3}-2}}{y^{-5/6}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{y^{-\frac{1}{3}}}{y^{-5/6}} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} y^{-\frac{1}{3}-\left(-\frac{5}{6} \right)} \\\\= y^{-\frac{1}{3}+\frac{5}{6}} .\end{array} To simplify the expression $-\dfrac{1}{3}+\dfrac{5}{6} ,$ change the expressions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $3$ and $6$ is $6$ since it is the lowest number that can be exactly divided by the denominators. Multiplying the terms by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to \begin{array}{l}\require{cancel} -\dfrac{1}{3}\cdot\dfrac{2}{2}+\dfrac{5}{6} \\\\= -\dfrac{2}{6}+\dfrac{5}{6} \\\\= \dfrac{-2+5}{6} \\\\= \dfrac{3}{6} \\\\= \dfrac{1}{2} .\end{array} The expression, $y^{-\frac{1}{3}+\frac{5}{6}} ,$ is equivalent to \begin{array}{l}\require{cancel} y^{\frac{1}{2}} \\\\= y^{1/2} .\end{array}