The student Richard solved a very simple exponential equation: $$ 2 \cdot 3^x=6 $$ in the following way.
(1) He transformed the left side and obtained: $$ 6^x=6 $$
(2) From the equality of the bases, he deduced: $$ x=1 $$ Did he make a mistake somewhere? If so, specify where.
Yes. The mistake is in step (1). In general, the equation $a \cdot b^n=(a \cdot b)^n$ does not hold.
Therefore, it is not possible to write $2 \cdot 3^x=(2 \cdot 3)^x=6^x$.
No. The calculation is correct.
Yes. The check is missing. The check is an integral part of the solution.
Yes. The mistake is in step (2). The student did not notice that the equation $6^x=6$ has two solutions, $x=1$ and $x=0$.
We show the correct solution to the equation: $$2 \cdot 3^x=6$$
We divide the equation by $2$ and obtain: $$ 3^x=3 $$ From the equality of the bases, we deduce the equality of the exponents: $$ x=1 $$ Note: The check is not necessary. In this case, all the transformations done on the equations are equivalent and do not change the solution set.