The teacher tasked his students with solving a logarithmic equation. One of the students was asked to present the solution on the board. The class observed the student’s work and was to assess the correctness of his solving procedure. The equation is: $$ \log_x 8=-3 $$
The student solved the equation in the following steps:
(1) First, he determined the condition for the base of the logarithm: $$ x\neq 1 \wedge x>0 $$ and wrote the domain of the equation: $$ (0,+\infty)\setminus {1} $$
(2) Next, he used the logarithmic identity: $$ \log_a x=v\Leftrightarrow x=a^v $$ to simplify the equation and obtained: $$ 8=x^{-3} $$
(3) He expressed $8$ as $2^3$ and rewrote the equation as: $$ 2^3=x^{-3} $$
(4) Then, he solved the equation in the following way: $$ \begin{aligned} 2^3=x^{-3} \cr 2^3=-x^3 \cr x=-2 \end{aligned} $$ Since the number $-2$ does not belong to the domain of the equation, the student concluded that the given equation has no solution.
Did the student make a mistake? If yes, identify in which step.
Yes. The mistake is in step (1) in the condition for the base of the logarithm. The domain of the equation should be $R\setminus {1}$.
Yes. The mistake is in step (2). It should be $(-3)^x=8$.
Yes. The mistake is in step (3). It is not possible to write the number $8$ in the form of $2^3$ because there is no power of $2$ on the right side of the equation.
Yes. The mistake is in step (4). Student solved the equation $2^3=x^{-3}$ incorrectly.
No. The whole procedure is correct.
Steps (1)-(3) are correct. The mistake is in the step (4). The equation from step (4) can be correctly solved as follows: $$ \begin{aligned} 2^3=x^{-3} \cr 2^3=[(x)^{-1} ]^3 \cr 2=(x)^{-1} \cr 2=\frac{1}{x} \cr x=\frac12 \end{aligned} $$ The root $x=\frac12$ belongs to the domain of the equation since it meets the condition for the base of the logarithm. The equation has only one solution: $$ x=\frac12 $$ In this case, the check is not necessary. However, we can perform it. $$ \begin{aligned} L & = \log_{\frac12} 8= \log_{\frac12} 2^3 =3 \log_{\frac12} 2=3 \log_{\frac12}\left(\frac12\right)^{-1}=-3 \log_{\frac12} \frac12 =-3 \cr R & =-3 \cr L & =R \end{aligned} $$