$ \log_2⁡(2 x)+\log_2⁡x=\log_2⁡3 $

Three students, Lukas, Magda, and Marie, were tested. Their task was to solve the following logarithmic equation on the board: $$ \log_2⁡(2 x)+\log_2⁡x=\log_2⁡3 $$ Initially, all three students identified the existence conditions for logarithms and determined the domain of the equation: $$ \begin{gather} 2x>0 \wedge x>0 \cr x\in(0;\infty) \end{gather} $$ Then they proceeded differently.

Lukas:

Lukas modified the left side of the equation as follows: $$\log_2⁡(3 x)=\log_2⁡3$$ Then, he stated that an equality of logarithms with the same base implies the equality of expressions within the logarithms. So, he obtained the following linear equation and solved it: $$ \begin{aligned} 3x & =3 \cr x & =1 \end{aligned} $$ Finally, he checked that the number $1$ belongs to the domain and concluded that $x=1$ is a solution.

Magda:

Magda also modified the left side of the equation to: $$\log_2⁡(2 x^2)=\log_2⁡3$$ She also believed that an equality of logarithms with the same base implies the equality of expressions within the logarithms. Therefore, she got the following quadratic equation and solved it: $$ \begin{aligned} 2x^2 & =3 \cr x^2 & =\frac32 \cr x_1 & =-\sqrt{\frac32};x_2=\sqrt{\frac32} \end{aligned} $$ Then, she observed that the number $-\sqrt{\frac32}$ does not belong to the domain of the equation. Therefore, Magda was convinced that the solution to the equation is only $x=\sqrt{\frac32}$.

Marie:

Marie also modified the left side of the equation using logarithmic rules to simplifying it. She wrote the following procedure on the board: $$ \begin{aligned} \log_2⁡2 x^2 & =\log_2⁡3 \cr 2 \log_2⁡2 x & =\log_2⁡3 \cr 2 (\log_2⁡2+\log_2⁡x) & =\log_2⁡3 \cr 2 \log_2⁡2+2\log_2⁡x & =\log_2⁡3 \cr 2+2\log_2⁡x & =\log_2⁡3 \cr 2\log_2⁡x & =8 – 2 \cr \log_2⁡x & =3 \cr x & =8 \end{aligned} $$ Finally, she checked that $8$ belongs to the domain of the equation and stated that $x=8$ is the solution.

Did any of them solve the equation correctly? If so, who?