## College Algebra (11th Edition)

$x=-\dfrac{1}{3}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\left( \dfrac{1}{3} \right)^{x+1}=9^x ,$ use the laws of exponents to express both sides in the same base. Then equate the exponents. $\bf{\text{Solution Details:}}$ 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 equation above is equivalent to \begin{array}{l}\require{cancel} \left( \dfrac{3}{1} \right)^{-(x+1)}=9^x \\\\ 3^{-(x+1)}=9^x .\end{array} Expressing both sides of the equation in the same base results to \begin{array}{l}\require{cancel} 3^{-(x+1)}=(3^2)^x \\\\ 3^{-(x+1)}=3^{2x} .\end{array} Since the bases are the same, then the exponents can be equated. That is, \begin{array}{l}\require{cancel} -(x+1)=2x \\\\ -x-1=2x \\\\ -1=2x+x \\\\ -1=3x \\\\ -\dfrac{1}{3}=x \\\\ x=-\dfrac{1}{3} .\end{array}