#### Answer

$x=-\dfrac{1}{3}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
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}