A teacher tasked the class with solving a logarithmic equation. Peter volunteered to perform the solution on the board. The class observed Peter's solution and stated that his procedure was not correct. The logarithmic equation was: $$ \log_3(x-1)+1= \log_3x $$
(1) First, Peter determined the existence condition for both logarithms, which specifies also the domain of the equation: $$ \begin{gather} x-1>0 \wedge x>0 \cr x \in (1;\infty) \end{gather} $$
(2) Next, applying the logarithm rules, he modified the left side of the equation: $$ \log_3( x-1+1)= \log_3x $$
(3) Then, by simplifying the equation, he obtained: $$ \begin{gather} \log_3x= \log_3x \cr 0=0 \end{gather} $$ which means that every number from the interval $(0;\infty)$ is a solution to the equation: $\log_3x=\log_3x$.
(4) Consequently, Peter concluded that every number from the domain of the given equation is a solution. So, the interval $(1;\infty)$ is the solution set of the equation $\log_3(x-1)+1= \log_3x$.
Where did Peter make a mistake?
The mistake is in step (1) in the existence condition. It should be: $$ x-1 \geq 0 \wedge x \geq 0 $$
The mistake is in step (2). The modification is incorrect.
The mistake is in step (3). It is not possible to obtain the same expression on both sides of the equation.
The mistake is in step (3). The equation in this step has no solution.
The existence condition for both logarithms was identified correctly. Petr made a mistake in step (2). Let us show the correct solution: $$ \begin{aligned} \log_3(x-1)+1 & = \log_3x \cr \log_3(x-1)+ \log_33 & = \log_3x \cr \log_3 (3(x-1)) & = \log_3x \cr \log_3(3x-3) & = \log_3x \cr 3x-3 & = x \cr 2x & = 3 \cr x & = \frac32 \end{aligned} $$ The root $x=\frac32$ belongs to the domain of the equation and so the equation has one unique solution. We can (but do not have to) do the check: $$ \begin{aligned} L &= \log_3 \left( \frac32-1\right )+1= \log_3 \frac12+1= \log_3 \frac12+ \log_33= \log_3 \frac32 \cr R &= \log_3 \frac32 \cr L &= R \end{aligned} $$