Find the mistake in the procedure for solving the given exponential equation: $$ 3^{2x}-12\cdot 3^x+27=0 $$
(1) First, the left side was modified by rewriting $3^{2x}$ as $3^2 3^x$: $$ 3^2 3^x-12\cdot 3^x+27=0 $$
(2) Then, the substitution was suggested: $$ 3^x=t $$ The equation was rewritten using the substitution variable $t$: $$ 3^2 t-12t+27=0 $$
(3) Next, the obtained linear equation was solved: $$ \begin{aligned} 9t-12t & = -27 \cr -3t & = -27 \cr t & = 9 \cr \end{aligned} $$
(4) Finally, from the used substitution, the solution for $x$ was found: $$ \begin{aligned} 3^x & = t \cr 3^x & = 9 \cr 3^x & = 3^2 \cr x & = 2 \end{aligned} $$
(5) Moreover, verification of the correctness of the solution was performed: $$ \begin{aligned} L & = 3^4-12\cdot 3^2+27=0 \cr R & = 0 \cr L & = R \end{aligned} $$ Is there a mistake in any step of the solving procedure? If so, specify where.
The whole procedure is correct.
The mistake is in step (1). The equality $3^{2x}=3^2 3^x$ does not generally hold.
The mistake is in step (2). It should have been $3t^2-12t+27=0$.
The mistake is in step (3). It should have been $t=3$.
The mistake is in step (4). The number $9$ should be written as $9=3^3$.
We show the correct procedure for solving the given equation: $$ 3^{2x}-12 \cdot 3^x+27=0 $$ It holds that: $$ 3^{2x}=(3^x )^2 $$ and the equation can be modified to: $$ (3^x )^2-12 \cdot 3^x+27=0 $$
Applying the substitution $3^x=t$, we obtain quadratic equation: $$ t^2-12t+27=0 $$ which has two solutions: $$ \begin{aligned} t_{1,2}=&\frac{12\pm \sqrt{144-4\cdot 27}}{2} \cr t_1=&\frac{12+6}{2}=9 \cr t_2=&\frac{12-6}{2}=3 \end{aligned} $$ Returning to the substitution $3^x=t$, we obtain: $$ \begin{aligned} 3^x=9 ~\lor~ & 3^x=3 \cr x=2 ~\lor~ & x=1 \end{aligned} $$ The equation has two solutions: $x=2$ and $x=1$. The check is not necessary.