$ \log_2(2x)-\log_2 8=1 $

Project ID: 
3000020098
SubArea: 
Logarithmic equations and inequalities
Question: 

Jane was tasked with solving a logarithmic equation: $$ \log_2(2x)-\log_2 8=1 $$ She solved it in the following way:

(1) First, she determined the domain of the logarithm: $$ \begin{aligned} 2x>0 \cr x \in (0;\infty) \end{aligned} $$

(2) Applying the logarithm rules, she modified the left side of the equation: $$ \log_2⁡2+\log_2⁡x-\log_2⁡2+\log_2⁡4=1 $$

(3) Then, she simplified the left side of the equation (subtracting $\log_2⁡2$): $$ \log_2⁡x+\log_2⁡4=1 $$

(4) She then modified the left side of the equation again to obtain the logarithmic equation in basic form: $$ \begin{aligned} \log_2⁡x+\log_2⁡2+\log_2⁡2=1 \cr \log_2⁡x+1+1=1 \cr \log_2⁡x=-1 \end{aligned} $$

(5) She applied the logarithm property: $$ \log_a⁡x=v \Leftrightarrow x=a^v $$ and received: $$ \begin{aligned} x & =2^{-1} \cr x & =\frac12 \end{aligned} $$

(6) She observed that this result belongs to the domain. Finally, she did the check: $$ \begin{aligned} L & =\log_2⁡(2 \cdot \frac12)-\log_2⁡8=\log_2⁡ \frac18=-3 \cr R & =1 \cr L & \neq R \end{aligned} $$

(7) Since the check did not come out, Jane stated that the given equation has no solution.

Did Jane make a mistake? If yes, identify where:

Answer 1: 

She made the mistake in step (1) regarding the domain. It should have been $2x\geq 0$.

Answer 2: 

She made the mistake in step (2). The modification of the left side of the equation is not correct.

Answer 3: 

She made the mistake in step (4). It is not true that $\log_2⁡4=\log_2⁡2+ \log_2⁡2$ .

Answer 4: 

The mistake is in step (5). It should have been: $$ \begin{aligned} \log_2⁡x & =-1 \cr x & =2^{-1}\cr x & =-2 \end{aligned} $$

Answer 5: 

There is no mistake in the procedure.

Fixed Answer: 
All Fixed
Correct Answer: 
Answer 2
Hint: 

Jane made the mistake in step (2) by forgetting to use parentheses. Instead of: $$ \log_2 2 + \log_2 x − \log_2 2 + \log_2 4 = 1 $$ it should have been: $$ \log_2 2 + \log_2 x − (\log_2 2 + \log_2 4) = 1 $$ Simplifying the previous equation, we obtain (realizing that $\log_2 4 = 2$): $$\begin{gather} \log_2 x = 3 \cr x = 2^3 \cr x = 8 \end{gather} $$ The root $x = 8$ belongs to the domain, and the equation has only one solution $x = 8$. We can, but do not have to, do the check.

Note: The equation can also be solved in another way. If we realize that $\log_2 8 = 3$ than we can simplify and solve the equation as follows: $$ \begin{gather} \log_2(2x) − \log_2 8 = 1 \cr \log_2(2x) = 4 \cr 2x = 2^4 \cr 2x = 16 \cr x = 8\end{gather} $$