Petr was examined in front of his classmates and was assigned to solve a logarithmic equation: $$ \log_2(x-2)+\log_2( x)=3 $$
His classmates were then supposed to assess the correctness of his solving procedure.
(1) Firstly, Petr determined the domains of both logarithms: $$ x-2>0 \land x>0 $$ By solving these conditions, he obtained the domain of the given equation: $$ (2,+\infty) $$
(2) He modified the left side of the equation according to the laws of logarithms: $$ \begin{align} \log_2( x-2+x)&=3 \cr \log_2( 2x-2)&=3 \end{align} $$
(3) Then, he used the rule: $$ \log_ax=v \Leftrightarrow x=a^v $$ and obtained: $$ 2x-2=2^3 $$
(4) Finally, he solved the linear equation: $$ \begin{align} 2x-2&=2^3 \cr 2x-2&=8 \cr 2x&=10 \cr x&=5 \end{align} $$
Petr observed that the root $x=5$ belongs to the domain of the equation.
(5) However, after doing the check: $$ \begin{align} L(5)&=\log_2(5-2)+\log_2( 5)=\log_23+\log_25\approx 3.907 \cr R(5) & =3 \cr L(5) & \neq R(5) \end{align} $$ he declared that the equation has no solution.
Did Peter make a mistake? If yes, identify where.
Yes. The mistake is in step (2). In general, the equality $$ \log_2 (x-2)+\log_2 x=\log_2 (2x-2) $$ does not hold.
Yes. The mistake is in step (1). The conditions for the domain of the given equation should be: $$ x-2 \geq 0 \land x\geq0 $$ Therefore, the domain of the equation is the interval $[ 2,+\infty)$. All of Peter's subsequent steps and the result are already correct.
Yes. The mistake is in step (5). The check is incorrect. It should be $$\log_23+\log_25=\log_28=3,$$ and therefore $x=5$ is the correct solution of the given equation.
No. All steps are correct.
Correct solution: $$\log_2(x-2)+\log_2( x)=3 $$ (1) We determine the domains of both logarithms: $$ x-2>0 \land x>0 $$ $$ x∈(2;\infty) $$
(2) We modify the left side of the equation according to the rule: $$ \log_a x+\log_a y=\log_a (xy) $$ and obtain: $$ \begin{align} \log_2[(x-2)\cdot x]=3 \cr \log_2( x^2-2x)=3 \end{align} $$
(3) Then, we simplify the equation according to the rule: $$ \log_ax=v\Leftrightarrow x=a^v $$ and get: $$ x^2-2x=2^3 $$
(4) And solve the quadratic equation: $$ \begin{align} x^2-2x-8&=0 \cr x_{1,2}&=\frac{2\pm \sqrt{36}}{2} \cr x_1&=4 \cr x_2&=-2 \end{align} $$
(5) The root $x=-2$ does not belong to the domain, so the equation has only a single solution $x = 4$. We may, but do not have to, perform a check. $$ \begin{align} L(4)&=\log_2(4-2)+\log_2( 4)=\log_22+\log_24=3 \cr R(4)&=3 \cr L(4)&=R(4) \end{align} $$