## College Algebra (11th Edition)

$x=-3$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\left( \sqrt[3]{5}\right)^{-x}=\left(\dfrac{1}{5}\right)^{x+2} ,$ express both sides of the equation in the same base. Once the bases are the same, equate the exponents. Use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \left( 5^{\frac{1}{3}} \right)^{-x}=\left(\dfrac{1}{5}\right)^{x+2} .\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} \left( 5^{\frac{1}{3}} \right)^{-x}=\left( 5^{-1}\right)^{x+2} .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} 5^{\frac{1}{3}(-x)}=5^{-1(x+2)} \\\\ 5^{-\frac{1}{3}x}=5^{-x-2} .\end{array} Since the bases are the same, then the exponents can be equated. Hence, \begin{array}{l}\require{cancel} -\frac{1}{3}x=-x-2 .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 3\left( -\frac{1}{3}x \right)=(-x-2)3 \\\\ -x=-3x-6 \\\\ -x+3x=-6 \\\\ 2x=-6 \\\\ x=-\dfrac{6}{2} \\\\ x=-3 .\end{array}