[Algebra]
Problem: Determine all real values of $x$ for which $3^{(x-1)} 9^{\frac{3}{2 x^{2}}}=27$.

Solution: Using the facts that $9=3^{2}$ and $27=3^{3}$, and the laws for manipulating exponents, we have

$$
\begin{aligned}
3^{x-1} 9^{\frac{3}{2 x^{2}}} & =27 \\
3^{x-1}\left(3^{2}\right)^{\frac{3}{2 x^{2}}} & =3^{3} \\
3^{x-1} 3^{\frac{3}{x^{2}}} & =3^{3} \\
3^{x-1+\frac{3}{x^{2}}} & =3^{3}
\end{aligned}
$$

When two powers of 3 are equal, their exponents must be equal so

$$
\begin{aligned}
x-1+\frac{3}{x^{2}} & =3 \\
x^{3}-x^{2}+3 & \left.=3 x^{2} \quad \text { (multiplying by } x^{2}\right) \\
x^{3}-4 x^{2}+3 & =0
\end{aligned}
$$

Since $x=1$ satisfies the equation, then $x-1$ is a factor of the left side. Using long division or synthetic division, we can factor this out to get $(x-1)\left(x^{2}-3 x-3\right)=0$.

Using the quadratic formula, the quadratic equation $x^{2}-3 x-3=0$ has roots

$$
x=\frac{3 \pm \sqrt{(-3)^{2}-4(1)(-3)}}{2}=\frac{3 \pm \sqrt{21}}{2}
$$

Therefore, the solutions to the original equation are $x=1$ and $x=\frac{3 \pm \sqrt{21}}{2}$.

Final answer: $1$,$\frac{3 + \sqrt{21}}{2}$,$\frac{3 - \sqrt{21}}{2}$