Answer
The solution set is $\left\{-\frac{1}{3}, -1\right\}$.
Work Step by Step
To solve the given equation, make the two sides have the same base.
Note that $9=3^2$ and $27=3^3$, so the given equation is equivalent to:
$(3^2)^{2x} \cdot (3^3)^{x^2} = 3^{-1}$
Use the rule $(a^m)^n = a^{mn}$ to obtain:
$3^{2(2x)} \cdot 3^{3(x^2)} = 3^{-1}
\\3^{4x} \cdot 3^{3x^2}=3^{-1}$
Use the rule $a^m \cdot a^n = a^{m+n}$ to obtain:
$3^{4x+3x^2} = 3^{-1}
\\3^{3x^2+4x} = 3^{-1}$
Use the rule $a^m=a^n \longrightarrow m=n$ to obtain:
$3x^2+4x=-1$
Add $1$ to both sides of the equation to obtain:
$\begin{array}{ccc}
&3x^2+4x +1 &= &-1+1
\\&3x^2+4x+1 &= &0
\end{array}$
Factor the trinomial to obtain:
$(3x+1)(x+1)=0$
Equate each factor to zero, and then solve each equation to obtain:
$\begin{array}{ccc}
&3x+1=0 &\text{ or } &x+1-0
\\&3x=-1 &\text{ or } &x=-1
\\&x=-\frac{1}{3} &\text{ or } &x=-1
\end{array}$
Thus, the solution set is $\left\{-\frac{1}{3}, -1\right\}$.
