Answer
$$
27^{x} =9^{x^{2+x}}
$$
The solution of the given equation is
$$
x=0 \text { or } \quad x=\frac{1}{2}.
$$
Work Step by Step
$$
27^{x} =9^{x^{2+x}}
$$
Since the bases must be the same, write 27as $3^{3 }$ and 9 as $ 3^{2} $ giving
$$
\begin{aligned} 27^{x} &=9^{x^{2+x}} \\\left(3^{3}\right)^{x} &=\left(3^{2}\right)^{x^{2+x}} \\ 3^{3 x} &=3^{2 x^{2}+2 x} \\ 3 x &=2 x^{2}+2 x \\ 0 &=2 x^{2}-x \\ 0 &=x(2 x-1) \\ x=0 & \text { or } \quad 2 x-1=0 \\ x=0 & \text { or } \quad x=\frac{1}{2} \end{aligned}
$$
So, the solution of the given equation is
$$
x=0 \text { or } \quad x=\frac{1}{2}.
$$
