## College Algebra (11th Edition)

$\dfrac{a^{11/8}}{b^{1/6}}$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $(a^{3/4}b^{2/3})(a^{5/8}b^{-5/6}) .$ $\bf{\text{Solution Details:}}$ 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} a^{\frac{3}{4}+\frac{5}{8}}b^{\frac{2}{3}+\left(-\frac{5}{6} \right)} \\\\= a^{\frac{3}{4}+\frac{5}{8}}b^{\frac{2}{3}-\frac{5}{6}} .\end{array} To add $\dfrac{3}{4}$ and $\dfrac{5}{8},$ change the fractions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $4$ and $8$ is $8$ since it is the lowest number that can be exactly divided by the denominators. Multiplying the fractions by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to \begin{array}{l}\require{cancel} \dfrac{3}{4}\cdot\dfrac{2}{2}+\dfrac{5}{8} \\\\= \dfrac{6}{8}+\dfrac{5}{8} \\\\= \dfrac{11}{8} .\end{array} To simplify the expression $\dfrac{2}{3}-\dfrac{5}{6} ,$ change the fractions 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 fractions by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to \begin{array}{l}\require{cancel} \dfrac{2}{3}\cdot\dfrac{2}{2}-\dfrac{5}{6} \\\\= \dfrac{4}{6}-\dfrac{5}{6} \\\\= \dfrac{4-5}{6} \\\\= \dfrac{-1}{6} \\\\= -\dfrac{1}{6} .\end{array} The expression, $a^{\frac{3}{4}+\frac{5}{8}}b^{\frac{2}{3}-\frac{5}{6}} ,$ simplifies to \begin{array}{l}\require{cancel} a^{\frac{11}{8}}b^{-\frac{1}{6}} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{a^{\frac{11}{8}}}{b^{\frac{1}{6}}} \\\\= \dfrac{a^{11/8}}{b^{1/6}} .\end{array}