Answer
$x=\dfrac{3}{7}$
Work Step by Step
Expressing both sides of the given equation, $
\left(\dfrac{1}{27}\right)^{x-1}=9^{2x}
,$ in the same base, the equation above is equivalent to
\begin{align*}
\left(\dfrac{1}{3^3}\right)^{x-1}&=\left(3^2\right)^{2x}
\\\\
\left(3^{-3}\right)^{x-1}&=\left(3^2\right)^{2x}
&(\text{use }\dfrac{1}{a^m}=a^{-m})
\\\\
3^{-3x+3}&=3^{4x}
&(\text{use }\left(a^m\right)^n=a^{mn})
.\end{align*}
Since $a^x=a^y$ implies $x=y$, the equation above implies
\begin{align*}
-3x+3&=4x
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
-3x+3x+3&=4x+3x
\\
3&=7x
\\\\
\dfrac{3}{7}&=\dfrac{\cancel7x}{\cancel7}
\\\\
\dfrac{3}{7}&=x
.\end{align*}
Hence, the solution set of the equation $
\left(\dfrac{1}{27}\right)^{x-1}=9^{2x}
$ is $x=\dfrac{3}{7}$.
